SSC Trigonometry Questions

Question 1 Show answer

If $$\tan B = \frac{5}{3}$$, what is the value of $$\frac{\cosec B + \sin B}{\cos B - \sec B}$$?

A

$$-\frac{177}{125}$$

B

$$-\frac{59}{15}$$

C

$$\frac{59}{15}$$

D

$$\frac{177}{125}$$

Question 2 Show answer

If $$6 \tan A \left(\tan A - 1\right) = 5 - \tan A$$, Given that O < A < $$\frac{\pi}{2}$$. what is the value of $$\left(\sin A + \cos A\right)$$?

A

$$3\sqrt{5}$$

B

$$\frac{3}{\sqrt{5}}$$

C

$$5\sqrt{3}$$

D

$$\frac{5}{\sqrt{3}}$$

Question 1 Show answer

A ladder leaning against a wall makes an angle $$\theta$$ with the horizontal ground such that $$\tan \theta = \frac{12}{5}.$$ If the height of the top of the ladder from the wall is 24 m, then what is the distance (in m) of the foot of the ladder from the wall?

A

18

B

19.5

C

7.5

D

10

Question 2 Show answer

$$\frac{\sin^2 52^\circ + 2 + \sin^{2} 38^\circ}{4 \cos^{2} 43^\circ - 5 + 4 \cos^{2} 47^\circ}$$ is:

A

3

B

$$\frac{1}{3}$$

C

-$$\frac{1}{3}$$

D

-3

Question 3 Show answer

ladder is resting against a wall. The angle between the foot of the ladder and wall is $$60^\circ$$, and the foot of the ladder is 3.6 m away from the wall. The length of the ladder (in m) is:

A

5.4

B

3.6

C

14.4

D

7.2

Question 4 Show answer

If $$4(cosec^2 57 - \tan^2 33) - \cos 90 + y * \tan^2 66 * \tan^2 24 = \frac{y}{2}$$, then the value of y is:

A

4

B

-4

C

8

D

-8

Question 5 Show answer

Let A and B be two towers with the same base. From the mid point of the line joining their feet, the angles of elevation of the tops of A and B are $$30^\circ$$ and $$45^\circ$$, respectively. The ratio of the heights of A and B is :

A

1 : 3

B

1: $$\sqrt{3}$$

C

$$\sqrt{3}$$: 1

D

3: 1

Question 6 Show answer

If $$4 \theta$$ is an acute angle , and $$\cot 4 \theta = \tan(\theta - 5^\circ)$$ is an acute angle, and $$\cot 46 = \tan(6 - 5^\circ)$$, then what is the value of $$\theta$$?

A

$$19^\circ$$

B

$$45^\circ$$

C

$$21^\circ$$

D

$$24^\circ$$

Question 7 Show answer

If $$4 - 2 \sin^2 \theta - 5 \cos \theta = 0, 0^\circ < \theta < 90^\circ$$, then the value of $$\cos \theta - \tan \theta$$ is:

A

$$\frac{1 + 2 \sqrt 3}{2}$$

B

$$\frac{2 - \sqrt 3}{2}$$

C

$$\frac{2 + \sqrt 3}{2}$$

D

$$\frac{1 - 2 \sqrt 3}{2}$$

Question 8 Show answer

Solve for $$ \theta: \cos^{2} - \sin^{2} \theta = \frac{1}{2}, 0 < \theta < 90^\circ$$.

A

$$45^\circ$$

B

$$60^\circ$$

C

$$30^\circ$$

D

$$40^\circ$$

Question 9 Show answer

If $$\cot \theta = \frac{1}{\sqrt{3}}, 0^\circ < \theta^\circ < 90^\circ$$ then the value of $$\frac{2 - \sin^{2} \theta}{1 - \cos^{2} \theta} + (\cosec^{2} \theta - \sec \theta)$$ is:

A

0

B

2

C

5

D

1

Question 10 Show answer

A person was standing on a road near a mall. He was 1215 m away from the mall and able to see the top of the mall from the road in such a way that the top of a tree, which is in between him and the mall, was exactly in line of sight with the top of the mall. The tree height is 20 m and it is 60 m away from him. How tall (in m) is the mall?

A

300

B

375

C

405

D

250

Question 11 Show answer

Let A and B be two towers with same base. From the midpoint of the line joining their feet. the angles of elevation of the tops of A and B are $$30^\circ$$ and $$60^\circ$$, respectively. The ratio of the heights of B and A is:

A

$$1 : 3$$

B

$$3 : 1$$

C

$$1 : 2$$

D

$$1 : \sqrt{3}$$

Question 12 Show answer

What is the average of sixty terms given below?
$$\cos^{2}x$$, $$\cos^{2}2x \cos^{2}3x$$,... $$\cos^{2}30x$$, $$\sin^{2}x$$, $$\sin^{2}2x$$, $$\sin^{2}3x$$,... $$ \sin^{2}30x$$

A

$$\cos^{2}x$$

B

0.5

C

1

D

$$\cos^{2}x\sin^{2}x$$

Question 13 Show answer

If $$3 \sin^2 \theta - \cos \theta - 1 = 0, 0^\circ < \theta < 90^\circ$$, then what is the value of $$\cot \theta + \cosec \theta ?$$

A

$$\frac{3\sqrt{2}}{2}$$

B

$$2\sqrt{3}$$

C

$$\sqrt{5}$$

D

$$2\sqrt{5}$$

Question 14 Show answer

If $$\sin A = \frac{1}{2}, A$$ is an acute angle, then find the value of $$\frac{\tan A - \cot A}{\sqrt{3}(1 + \cosec A)}$$

A

$$\frac{4\sqrt{3}}{9}$$

B

$$\frac{2}{9}$$

C

$$-\frac{2}{9}$$

D

$$-\frac{4\sqrt{3}}{9}$$

Question 15 Show answer

If $$\frac{\cos^2 \theta}{\cot^2 \theta + \sin^2 \theta - 1} = 3, 0^\circ < \theta < 90^\circ$$, then the value of $$(\tan \theta + \cosec \theta)$$ is:

A

$$\frac{5\sqrt{3}}{3}$$

B

$$3\sqrt{3}$$

C

$$2\sqrt{3}$$

D

$$\frac{4\sqrt{3}}{3}$$

Question 16 Show answer

$$1 + 2 \tan^2 \theta + 2 \sin \theta \sec^2 \theta, 0^\circ < \theta < 90^\circ$$, is equal to:

A

$$\frac{1 - \cos \theta}{1 + \cos \theta}$$

B

$$\frac{1 + \cos \theta}{1 - \cos \theta}$$

C

$$\frac{1 - \sin \theta}{1 + \sin \theta}$$

D

$$\frac{1 + \sin \theta}{1 - \sin \theta}$$

Question 17 Show answer

If $$\frac{\sin^2 \theta}{\tan^2 \theta - \sin^2 \theta} = 5, \theta$$ is an acute angle, then the value of $$\frac{24\sin^2\theta-15\sec^2\theta}{6\operatorname{cosec}^2\theta-7\cot^2\theta}$$ is:

A

-2

B

2

C

-14

D

14

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Solution

$$\frac{\sin^2\theta}{\tan^2\theta-\sin^2\theta}=5$$

$$\frac{\sin^2\theta}{\frac{\sin^2\theta\ }{\cos^2\theta\ }-\sin^2\theta}=5$$

$$\frac{\sin^2\theta}{\sin^2\theta\ \left(\frac{1\ }{\cos^2\theta\ }-1\right)}=5$$

$$\frac{\cos^2\theta}{1-\cos^2\theta\ }=5$$

$$\frac{\cos^2\theta}{\sin^2\theta\ \ }=5$$

$$\cot^2\theta\ \ =5$$

$$\operatorname{cosec}^2\theta\ =1+\cot^2\theta\ =1+5=6$$

$$\sin^2\theta=\frac{1}{\operatorname{cosec}^2\theta}\ =\frac{1}{6}$$

$$\cos^2\theta\ =1-\sin^2\theta=1-\frac{1}{6}=\frac{5}{6}$$

$$\sec^2\theta\ =\frac{1}{\cos^2\theta\ }=\frac{6}{5}$$

$$\frac{24\sin^2\theta-15\sec^2\theta}{6\operatorname{cosec}^2\theta-7\cot^2\theta}=\frac{24\left(\frac{1}{6}\right)-15\left(\frac{6}{5}\right)}{6\left(6\right)-7\left(5\right)}$$

$$=\frac{4-18}{36-35}$$

$$=-14$$

Hence, the correct answer is Option C

Question 18 Show answer

If $$\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}=5$$, then the value of $$\frac{4\sin^2\theta+3}{2\cos^2\theta+2}$$ is:

A

$$\frac{75}{17}$$

B

$$\frac{75}{34}$$

C

$$\frac{1}{2}$$

D

$$\frac{3}{2}$$

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Solution

$$\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}=5$$

$$\sin\theta+\cos\theta=5\sin\theta\ -5\cos\theta\ $$

$$4\sin\theta=6\cos\theta\ $$

$$\tan\theta\ =\frac{3}{2}$$

$$\sec\theta\ =\sqrt{\left(\frac{3}{2}\right)^2+1}=\frac{\sqrt{13}}{2}$$

$$\cos\theta\ =\frac{2}{\sqrt{13}}$$

$$\sin\theta\ =\sqrt{1-\left(\frac{2}{\sqrt{13}}\right)^2}=\frac{3}{\sqrt{13}}$$

$$\frac{4\sin^2\theta+3}{2\cos^2\theta+2}=\frac{4\left(\frac{3}{\sqrt{13}}\right)^2+3}{2\left(\frac{2}{\sqrt{13}}\right)^2+2}$$

$$=\frac{\frac{36}{13}+3}{\frac{8}{13}+2}$$

$$=\frac{\frac{36+39}{13}}{\frac{8+26}{13}}$$

$$=\frac{75}{34}$$

Hence, the correct answer is Option B

Question 19 Show answer

For $$0^\circ < \theta < 90^\circ, \frac{1}{\cos \theta} + \frac{1}{\tan \theta - \sec \theta}$$ is equal to:

A

$$\sec \theta$$

B

$$-\sec \theta$$

C

$$-\tan \theta$$

D

$$\tan \theta$$

Question 20 Show answer

If $$3 \tan \theta = 2\sqrt{3} \sin \theta, 0^\circ < \theta < 90^\circ$$, then find the value of $$2 \sin^2 2\theta - 3 \cos^2 3\theta$$.

A

1

B

$$\frac{3}{2}$$

C

$$\frac{1}{2}$$

D

$$-\frac{3}{2}$$

View Explanation Hide Explanation

Solution

$$3\tan\theta=2\sqrt{3}\sin\theta$$

$$3\frac{\sin\theta\ }{\cos\theta\ }=2\sqrt{3}\sin\theta$$

$$\cos\theta\ =\frac{3}{2\sqrt{3}}$$

$$\cos\theta\ =\frac{\sqrt{3}}{2}$$

$$\theta\ =30^{\circ\ }$$ [$$0^\circ < \theta < 90^\circ$$]

$$2\sin^22\theta-3\cos^23\theta=2\sin^260^{\circ\ }-3\cos^290^{\circ\ }$$

= $$2\left(\frac{\sqrt{3}}{2}\right)^2-3\left(0\right)^2$$

= $$\frac{3}{2}$$

Hence, the correct answer is Option B

Question 21 Show answer

If $$\sec(5 \alpha - 15^\circ) = \cosec(15^\circ - 2 \alpha)$$, then the value of $$\cos \alpha + \sin 2 \alpha + \tan(1.5 \alpha)$$ is:

A

$$\sqrt{3} + 1$$

B

$$\sqrt{2} - 1$$

C

$$\sqrt{3} - 1$$

D

$$\sqrt{2} + 1$$

Question 22 Show answer

If $$\sin^2 \theta = 2 \sin \theta - 1, 0^\circ \leq \theta \leq 90^\circ $$, then find the value of: $$\frac{1 + \cosec \theta}{1 - \cos \theta}$$.

A

-2

B

1

C

2

D

-1

View Explanation Hide Explanation

Solution

$$\sin^2\theta=2\sin\theta-1$$

$$\sin^2\theta-2\sin\theta+1=0$$

$$\left(\sin\theta-1\right)^2=0$$

$$\sin\theta-1=0$$

$$\sin\theta=1$$

$$0^{\circ}\le\theta\le90^{\circ}$$

$$\Rightarrow$$  $$\theta=90^{\circ}$$

$$\frac{1+\operatorname{cosec}\theta}{1-\cos\theta}=\frac{1+\operatorname{cosec}90^{\circ\ }}{1-\cos90^{\circ\ }}$$

$$=\frac{1+1\ }{1-0\ }$$

$$=2$$

Hence, the correct answer is Option C

Question 23 Show answer

If $$\tan \theta = \sqrt{5}$$, then the value of $$\frac{\cosec^2 \theta + \sec^2 \theta}{\cosec^2 \theta - \sec^2 \theta}$$ is:

A

$$-\frac{7}{5}$$

B

$$\frac{7}{5}$$

C

$$-\frac{3}{2}$$

D

$$\frac{3}{2}$$

Question 24 Show answer

If $$\sin \left(\frac{2A + B}{2}\right) = \cos \left(\frac{2A - B}{2}\right) = \frac{\sqrt{3}}{2}, 0^\circ < \frac{2A + B}{2} < 90^\circ$$ and $$0^\circ < \frac{2A + B}{2} < 90^\circ$$ then find the value of $$\sin[3(A - B)]$$.

A

$$\frac{\sqrt{3}}{2}$$

B

$$\frac{1}{\sqrt{2}}$$

C

$$\frac{1}{2}$$

D

1

Question 25 Show answer

If $$\sin(20 + x)^\circ = \cos 60^\circ, 0 \leq (20 + x) \leq 90$$, then find the value of $$2 \sin^2(3x + 15)^\circ - \cosec^2(2x + 10)^\circ$$.

A

3

B

-3

C

$$-\frac{1}{3}$$

D

-2

View Explanation Hide Explanation

Solution

$$\sin(20+x)^{\circ}=\cos60^{\circ}$$

$$\sin(20+x)^{\circ}=\sin30^{\circ}$$

$$20+x=30$$

$$x=10$$

$$2\sin^2(3x+15)^{\circ}-\operatorname{cosec}^2(2x+10)^{\circ}=2\sin^245^{\circ}-\operatorname{cosec}^230^{\circ}$$

$$=2\left(\frac{1}{\sqrt{2}}\right)^2-\left(2\right)^2$$

$$=2\left(\frac{1}{2}\right)-4$$

$$=1-4$$

$$=-3$$

Hence, the correct answer is Option B

Question 26 Show answer

If $$\sin^2 \theta - \cos^2 \theta - 3 \sin \theta + 2 = 0, 0^\circ < \theta < 90^\circ$$, then what is the value of $$\frac{1}{\sqrt{\sec \theta - \tan \theta}}$$ is:

A

$$\sqrt[4]{3}$$

B

$$\sqrt[2]{2}$$

C

$$\sqrt[2]{3}$$

D

$$\sqrt[4]{2}$$

View Explanation Hide Explanation

Solution

$$\sin^2\theta-\cos^2\theta-3\sin\theta+2=0$$

$$\sin^2\theta-\left(1-\sin^2\theta\ \right)-3\sin\theta+2=0$$

$$2\sin^2\theta-3\sin\theta+1=0$$

$$2\sin^2\theta-2\sin\theta-\sin\theta\ +1=0$$

$$2\sin\theta\ \left(\sin\theta\ -1\right)-1\left(\sin\theta\ -1\right)=0$$

$$\left(\sin\theta\ -1\right)\left(2\sin\theta\ -1\right)=0$$

$$\sin\theta\ -1=0$$  or  $$2\sin\theta\ -1=0$$

$$\sin\theta\ =1$$  or  $$\sin\theta\ =\frac{1}{2}$$

$$\theta\ =90^{\circ\ }$$  or  $$\theta\ =30^{\circ\ }$$

Given, $$0^\circ < \theta < 90^\circ$$

$$\Rightarrow$$  $$\theta\ =30^{\circ\ }$$

$$\frac{1}{\sqrt{\sec\theta-\tan\theta}}=\frac{1}{\sqrt{\sec30^{\circ\ }-\tan30^{\circ\ }}}$$

$$=\frac{1}{\sqrt{\frac{2}{\sqrt{3}}\ -\frac{1}{\sqrt{3}}}}$$

$$=\frac{1}{\sqrt{\frac{1}{\sqrt{3}}}}$$

$$=\sqrt[\ 4]{3}$$

Hence, the correct answer is Option A

Question 27 Show answer

Find the value of $$\frac{3}{4} \cot^2 30^\circ + \cos^2 30^\circ - 3\cosec^2 60^\circ + \tan^2 60^\circ$$

A

-4

B

10

C

$$\frac{\sqrt{3}}{4}$$

D

2

Question 28 Show answer

If $$2 \cos^2 \theta - 5 \cos \theta + 2 = 0, 0^\circ < \theta < 90^\circ$$, then the value of $$(\sec \theta + \tan \theta)$$ is:

A

$$2 + \sqrt{3}$$

B

$$1 - \sqrt{3}$$

C

$$1 + \sqrt{3}$$

D

$$2 - \sqrt{3}$$

View Explanation Hide Explanation

Solution

$$2\cos^2\theta-5\cos\theta+2=0$$

$$2\cos^2\theta-4\cos\theta-\cos\theta+2=0$$

$$2\cos\theta\left(\cos\theta-2\right)-1\left(\cos\theta-2\right)=0$$

$$\left(\cos\theta-2\right)\left(2\cos\theta-1\right)=0$$

$$\cos\theta=2$$  or $$\cos\theta=\frac{1}{2}$$

$$\cos\theta=2$$ is not possible.

So, $$\cos\theta=\frac{1}{2}$$

$$\sec\theta=2$$

$$\tan\theta=\sqrt{\sec^2\theta-1}=\sqrt{4-1}=\sqrt{3}$$

$$\therefore$$  $$\left(\sec\theta+\tan\theta\right)=2+\sqrt{3}$$

Hence, the correct answer is Option A

Question 29 Show answer

If $$\frac{1}{1 - \sin \theta} + \frac{1}{1 + \sin \theta} = 4 \sec \theta, 0^\circ < \theta < 90^\circ$$, then the value of $$\cot\theta+\operatorname{cosec}\theta$$ is:

A

$$\frac{4\sqrt{3}}{3}$$

B

$$\sqrt{3}$$

C

$$\frac{5\sqrt{3}}{3}$$

D

$$3\sqrt{3}$$

View Explanation Hide Explanation

Solution

$$\frac{1}{1-\sin\theta}+\frac{1}{1+\sin\theta}=4\sec\theta$$

$$\frac{1+\sin\theta\ +1-\sin\theta\ }{1-\sin^2\theta}=\frac{4}{\cos\theta}$$

$$\frac{2}{\cos^2\theta}=\frac{4}{\cos\theta}$$

$$\cos\theta=\frac{1}{2}$$

$$0^\circ < \theta < 90^\circ$$

$$\Rightarrow$$  $$\theta=60^{\circ}$$

$$\cot\theta+\operatorname{cosec}\theta=\cot60^{\circ}+\operatorname{cosec}60^{\circ}$$

= $$\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}$$

= $$\frac{3}{\sqrt{3}}$$

= $$\sqrt{3}$$

Hence, the correct answer is Option B

Question 30 Show answer

If $$\tan \theta + 3 \cot \theta - 2\sqrt{3} = 0, 0^\circ < \theta < 90^\circ$$, then what is the value of $$(\cosec^2 \theta + \cos^2 \theta)?$$

A

$$\frac{2}{3}$$

B

$$\frac{19}{12}$$

C

$$\frac{14}{3}$$

D

$$\frac{11}{12}$$

View Explanation Hide Explanation

Solution

$$\tan\theta+3\cot\theta-2\sqrt{3}=0$$

$$\tan\theta+\frac{3}{\tan\theta\ }-2\sqrt{3}=0$$

$$tan^2\theta-2\sqrt{3}\tan\theta\ +3=0$$

$$\left(\tan\theta\ -\sqrt{3}\right)^2=0$$

$$\tan\theta\ -\sqrt{3}=0$$

$$\tan\theta\ =\sqrt{3}$$

$$0^\circ < \theta < 90^\circ$$

$$\Rightarrow$$  $$\theta\ =60^{\circ\ }$$

$$\operatorname{cosec}^2\theta+\cos^2\theta=\operatorname{cosec}^260^{\circ\ }+\cos^260^{\circ\ }$$

= $$\left(\frac{2}{\sqrt{3}}\right)^2\ +\left(\frac{1}{2}\right)^2$$

= $$\frac{4}{3}\ +\frac{1}{4}$$

= $$\frac{16+3}{12}$$

= $$\frac{19}{12}$$

Hence, the correct answer is Option B

Question 31 Show answer

Simplify: $$\frac{(\sin \theta + \sec \theta)^2 + (\cos \theta + \cosec \theta)^2}{(1 + \sec \theta \cosec \theta)^2}, 0^\circ < \theta < 90^\circ$$

A

0

B

1

C

-1

D

2

Question 32 Show answer

The value of $$\frac{\tan13^{\circ}\tan36^{\circ}\tan45^{\circ}\tan54^{\circ}\tan77^{\circ}}{2\sec^260^{\circ}(\sin^260^{\circ}-3\cos60^{\circ}+2)}$$ is:

A

$$\frac{1}{10}$$

B

$$-\frac{1}{4}$$

C

$$\frac{1}{4}$$

D

$$-\frac{1}{10}$$

View Explanation Hide Explanation

Solution

$$\frac{\tan13^{\circ}\tan36^{\circ}\tan45^{\circ}\tan54^{\circ}\tan77^{\circ}}{2\sec^260^{\circ}(\sin^260^{\circ}-3\cos60^{\circ}+2)}=\frac{\tan13^{\circ}\tan36^{\circ}\left(1\right)\tan\left(90-36^{\circ}\right)\tan\left(90-13^{\circ\ }\right)}{2\left(2\right)^2\left(\left(\frac{\sqrt{3}}{2}\right)^2-3\left(\frac{1}{2}\right)+2\right)}$$

$$=\frac{\tan13^{\circ}\tan36^{\circ}\cot36^{\circ}\cot13^{\circ\ }}{2\left(4\right)\left(\frac{3}{4}-\frac{3}{2}+2\right)}$$

$$=\frac{1}{8\left(\frac{3-6+8}{4}\right)}$$

$$=\frac{1}{2\left(5\right)}$$

$$=\frac{1}{10}$$

Hence, the correct answer is Option A

Question 33 Show answer

The value of
$$(\sin 37^\circ \cos 53^\circ + \cos 37^\circ \sin 53^\circ) - \frac{4 \cos^2 37^\circ - 7 + 4 \cos^2 53^\circ}{\tan^2 47^\circ + 4 - \cosec^2 43^\circ}$$ is:

A

-2

B

0

C

1

D

2

Question 34 Show answer

$$\frac{\cosec \theta}{\cosec \theta - 1} + \frac{\cosec \theta}{\cosec \theta + 1} - \tan^2 \theta, 0^\circ < \theta < 90^\circ$$, is equal to:

A

$$\sec^2 \theta + 1$$

B

$$\sec^2 \theta$$

C

$$2 \sec^2 \theta$$

D

$$1 - \tan^2 \theta$$

Question 35 Show answer

$$\frac{\cot^3 \theta}{\cosec^2 \theta} + \frac{\tan^3 \theta}{\sec^2 \theta} + 2\sin \theta \cos \theta = ?$$

A

$$\cosec \theta \sec \theta$$

B

$$\sin \theta \cos \theta$$

C

$$\cosec^2 \theta \sec^2 \theta$$

D

$$\sin^2 \theta \cos \theta$$

Question 36 Show answer

In $$\triangle$$ABC, right angled at B, if cot A = $$\frac{1}{2}$$, then the value of $$\frac{\sin A(\cos C + \cos A)}{\cos C(\sin C - \sin A)}$$ is

A

3

B

-3

C

-2

D

2

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Solution

cot A = $$\frac{\text{Adjacent side}}{\text{Opposite side}}$$ $$\frac{1}{2}$$

$$\frac{\sin A(\cos C + \cos A)}{\cos C(\sin C - \sin A)}$$ = $$\frac{\frac{2}{\sqrt{5}}\left(\frac{2}{\sqrt{5}}+\frac{1}{\sqrt{5}}\right)}{\frac{2}{\sqrt{5}}\left(\frac{1}{\sqrt{5}}-\frac{2}{\sqrt{5}}\right)}$$

= $$\frac{\left(\frac{3}{\sqrt{5}}\right)}{\left(-\frac{1}{\sqrt{5}}\right)}$$

= -3

Hence, the correct answer is Option B

Question 37 Show answer

The value of $$\frac{\sec^2 60^\circ \cos^2 45^\circ + \cosec^2 30^\circ}{\cot 30^\circ \sec^2 45^\circ - \cosec^2 30^\circ \tan 45^\circ}$$ is:

A

$$-3(2 + \sqrt{3})$$

B

$$3(2 - \sqrt{3})$$

C

$$-3(2 - \sqrt{3})$$

D

$$3(2 + \sqrt{3})$$

View Explanation Hide Explanation

Solution

$$\frac{\sec^260^{\circ}\cos^245^{\circ}+\operatorname{cosec}^230^{\circ}}{\cot30^{\circ}\sec^245^{\circ}-\operatorname{cosec}^230^{\circ}\tan45^{\circ}}=\frac{\left(2\right)^2.\left(\frac{1}{\sqrt{2}}\right)^2+\left(2\right)^2}{\left(\sqrt{3}\right)\left(\sqrt{2}\right)^2-\left(2\right)^2.\left(1\right)}$$

$$=\frac{4\times\frac{1}{2}+4}{2\sqrt{3}-4}$$

$$=\frac{6}{2\sqrt{3}-4}$$

$$=\frac{3}{\sqrt{3}-2}$$

$$=\frac{3}{\sqrt{3}-2}\times\frac{\sqrt{3}+2}{\sqrt{3}+2}$$

$$=\frac{3\left(\sqrt{3}+2\right)}{3-4}$$

$$=-3\left(2+\sqrt{3}\right)$$

Hence, the correct answer is Option A

Question 38 Show answer

Find the value of $$\tan35^{\circ}\cot40^{\circ}\tan45^{\circ}\cot50^{\circ}\tan55^{\circ}$$.

A

$$\frac{1}{\sqrt{2}}$$

B

$$\frac{1}{2}$$

C

-1

D

1

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Solution

$$\tan35^{\circ}\cot40^{\circ}\tan45^{\circ}\cot50^{\circ}\tan55^{\circ}=\tan35^{\circ}\cot40^{\circ}\tan45^{\circ}\cot\left(90-40^{\circ}\right)\tan\left(90-35^{\circ}\right)$$

$$=\tan35^{\circ}\cot40^{\circ}\left(1\right)\tan40^{\circ}\cot35^{\circ}$$

$$=1$$

Hence, the correct answer is Option D

Question 39 Show answer

If $$\sec 31^\circ = x,$$ then $$\sin^2 59^\circ + \frac{1}{\sec^2 31^\circ} - \frac{1}{\sec^2 59^\circ \cosec^2 59^\circ}$$ is equal to:

A

$$\frac{x^2 - 2}{x^2}$$

B

$$\frac{2 - x^2}{x}$$

C

$$\frac{2 - x^2}{x^2}$$

D

$$\frac{x^2 - 2}{x}$$

Question 40 Show answer

Find the value of $$\frac{\tan^2 30^\circ}{\sec^2 30^\circ} + \frac{\cosec^2 45^\circ}{\cot^2 45^\circ} - \frac{\sec^2 60^\circ}{\cosec^2 60^\circ}$$

A

$$-\frac{3}{4}$$

B

$$\frac{5}{4}$$

C

$$\frac{13}{4}$$

D

$$\frac{23}{12}$$

View Explanation Hide Explanation

Solution

$$\frac{\tan^230^{\circ}}{\sec^230^{\circ}}+\frac{\operatorname{cosec}^245^{\circ}}{\cot^245^{\circ}}-\frac{\sec^260^{\circ}}{\operatorname{cosec}^260^{\circ}}=\frac{\left(\frac{1}{\sqrt{3}}\right)^2}{\left(\frac{2}{\sqrt{3}}\right)^2}+\frac{\left(\sqrt{2}\right)^2}{\left(1\right)^2}-\frac{\left(2\right)^2}{\left(\frac{2}{\sqrt{3}}\right)^2}$$

$$=\frac{\frac{1}{3}}{\frac{4}{3}}+\frac{2}{1}-\frac{4}{\frac{4}{3}}$$

$$=\frac{1}{4}+2-\frac{4\times3}{4}$$

$$=\frac{-3}{4}$$

Hence, the correct answer is Option A

Question 41 Show answer

Find the value of $$\operatorname{cosec}(60^{\circ}+A)-\sec(30^{\circ}-A)+\frac{\operatorname{cosec}49^{\circ}}{\sec41^{\circ}}$$.

A

1

B

0

C

-1

D

2

View Explanation Hide Explanation

Solution

$$\operatorname{cosec}(60^{\circ}+A)-\sec(30^{\circ}-A)+\frac{\operatorname{cosec}49^{\circ}}{\sec41^{\circ}}$$

= $$\operatorname{cosec}(60^{\circ}+A)-\sec(90^{\circ}-60^{\circ}-A)+\frac{\operatorname{cosec}49^{\circ}}{\sec\left(90-49\right)^{\circ}}$$

$$\left[\sec\left(90\ -\theta\right)=\operatorname{cosec}\theta\right]$$

= $$\operatorname{cosec}\left(60^{\circ}+A\right)-\sec\left(90^{\circ}-\left(60^{\circ}+A\right)\right)+\frac{\operatorname{cosec}49^{\circ}}{\operatorname{cosec}49^{\circ}}$$

= $$\operatorname{cosec}\left(60^{\circ}+A\right)-\operatorname{cosec}\left(60^{\circ}+A\right)+1$$

= 1

Hence, the correct answer is Option A

Question 42 Show answer

Find the value of $$\frac{8 \sin 30^\circ \sin^2 60^\circ - 4 \sin 90^\circ - \sec^2 45^\circ}{\tan^2 45^\circ - \cot^2 30^\circ}$$.

A

$$-\frac{1}{2}$$

B

$$\frac{5}{2}$$

C

$$\frac{3}{4}$$

D

$$\frac{3}{2}$$

Question 43 Show answer

If $$\cot \theta = \frac{15}{8}, \theta$$ is an acute angle, then find the value of $$\frac{(1 - \cos \theta)(2 + 2 \cos \theta)}{(2 - 2 \sin \theta)(1 + \sin \theta)}$$.

A

$$\frac{225}{64}$$

B

$$\frac{64}{225}$$

C

$$\frac{16}{15}$$

D

$$\frac{8}{15}$$

View Explanation Hide Explanation

Solution

$$\cot\theta=\frac{15}{8}$$

$$\tan\theta=\frac{8}{15}$$

$$\sec\theta=\sqrt{1+\tan^2\theta\ }=\sqrt{1+\left(\frac{8}{15}\right)^2}=\sqrt{\frac{289}{225}\ }=\frac{17}{15}$$

$$\cos\theta=\frac{15}{17}$$

$$\sin\theta=\sqrt{1-\cos^2\theta\ }=\sqrt{1-\left(\frac{15}{17}\right)^2}=\sqrt{1-\frac{225}{289}}=\sqrt{\frac{64}{289}}=\frac{8}{17}$$

$$\frac{(1-\cos\theta)(2+2\cos\theta)}{(2-2\sin\theta)(1+\sin\theta)}=\frac{\left(1-\frac{15}{17}\right)\left[2+2\left(\frac{15}{17}\right)\right]}{\left[2-2\left(\frac{8}{17}\right)\right]\left(1+\frac{8}{17}\right)}$$

$$=\frac{\left(\frac{2}{17}\right)\left[\frac{64}{17}\right]}{\left[\frac{18}{17}\right]\left(\frac{25}{17}\right)}$$

$$=\frac{2\times64}{18\times25}$$

$$=\frac{64}{225}$$

Hence, the correct answer is Option B

Question 44 Show answer

If $$\frac{\cosec \theta + \cot \theta}{\cosec \theta - \cot \theta} = 7$$, then the value of $$\frac{4 \sin^2 \theta - 1}{4 \sin^2 \theta + 5}$$ is:

A

$$-\frac{1}{9}$$

B

$$\frac{1}{3}$$

C

$$-\frac{1}{3}$$

D

$$\frac{1}{9}$$

Question 45 Show answer

$$(\sqrt{\sec^2 \theta + \cosec^2 \theta})\left(\frac{\sin \theta(1 + \cos \theta)}{1 + \cos \theta - \sin^2 \theta}\right), 0^\circ < \theta < 90^\circ$$ is equal to:

A

$$\sec^2 \theta$$

B

$$\cosec^2 \theta$$

C

$$\cot \theta$$

D

$$\tan \theta$$

Question 46 Show answer

If $$7 \cos^2 \theta + 5 \sin^2 \theta - 6 = 0, (0^\circ < \theta < 90^\circ)$$, then what is the value of $$\sqrt{\frac{\cosec \theta + \tan \theta}{\sec \theta - \cot \theta}}$$

A

$$\sqrt{2} + 1$$

B

$$\sqrt{2} - 1$$

C

$$\sqrt{3} + 1$$

D

$$\sqrt{3} - 1$$

Question 47 Show answer

The value of $$4(\sin^4 30^\circ + \cos^4 30^\circ) - 3(\sin^2 45^\circ - 2 \cos^2 45^\circ)$$ is:

A

0

B

4

C

2

D

1

Question 48 Show answer

Find the value of $$\cot 25^\circ \cot 35^\circ \cot 45^\circ \cot 55^\circ \cot 65^\circ$$.

A

$$\frac{1}{\sqrt{3}}$$

B

$$\sqrt{3}$$

C

$$\frac{\sqrt{3}}{2}$$

D

1

Question 49 Show answer

If $$5 \sin^2 \theta - 4 \cos \theta - 4 = 0, 0^\circ < \theta < 90^\circ$$, then the value of $$(\cot \theta + \cosec \theta)$$ is:

A

$$\frac{\sqrt{6}}{2}$$

B

$$\frac{\sqrt{6}}{3}$$

C

$$\frac{3}{2}$$

D

$$\frac{2}{3}$$

Question 50 Show answer

If $$3 \cos^2 \theta - 4 \sin \theta + 1 = 0, 0^\circ < \theta < 90^\circ$$, then $$\tan \theta + \sec \theta = ?$$

A

$$2\sqrt{5}$$

B

$$2\sqrt{3}$$

C

$$3\sqrt{3}$$

D

$$\sqrt{5}$$

Question 51 Show answer

The value of
$$\frac{4 \sin^2 30^\circ \tan 60^\circ - 3 \cos^2 60^\circ \sec^2 30^\circ}{4 \cot^2 45^\circ - \sec^2 60^\circ + \sin^2 60^\circ + \cos^2 90^\circ}$$ is:

A

$$\frac{2(\sqrt{3} - 2)}{3}$$

B

$$\frac{1}{3}(\sqrt{3} + 2)$$

C

$$\frac{2(\sqrt{3} + 2)}{3}$$

D

$$\frac{1}{3}(\sqrt{3} - 2)$$

Question 52 Show answer

If $$\cos \theta - \sin \theta = \sqrt{3} \cos(90^\circ - \theta), 0^\circ < \theta < 90^\circ$$ then find the value of $$\tan \theta - cot \theta$$.

A

$$\frac{3 - 2\sqrt{3}}{(1 + \sqrt{3})}$$

B

$$-\frac{3 + 2\sqrt{3}}{(1 + \sqrt{3})}$$

C

$$-\frac{3 + 2\sqrt{3}}{(1 - \sqrt{3})}$$

D

$$\frac{3 + 2\sqrt{3}}{(1 + \sqrt{3})}$$

Question 53 Show answer

If $$\sin(A + B) = 1$$ and $$\cos(A - B) = \frac{\sqrt{3}}{2}, A + B \leq 90^\circ$$ and $$A > B$$, then the value of $$\frac{5 \sin^2 B + 4 \tan^2 A}{2 \sin B \cos A}$$ is:

A

$$16\frac{1}{2}$$

B

18

C

20

D

$$26\frac{1}{2}$$

Question 54 Show answer

If $$\cos(2 \theta + 54^\circ) = \sin \theta, 0^\circ < (2 \theta + 54^\circ) < 90^\circ$$, then what is the value of $$\frac{1}{\cot 5 \theta + \sec \frac{5 \theta}{2}}$$?

A

$$\frac{\sqrt{3}}{2}$$

B

$$\frac{2\sqrt{3}}{3}$$

C

$$\frac{\sqrt{3}}{3}$$

D

$$\frac{1}{3}$$

Question 55 Show answer

If $$\sin^6 \theta + \cos^6 \theta = \frac{1}{3}, 0^\circ < \theta < 90^\circ$$, then what is the value of $$\sin \theta \cos \theta$$?

A

$$\frac{\sqrt{2}}{3}$$

B

$$\frac{\sqrt{6}}{6}$$

C

$$\frac{\sqrt{2}}{\sqrt{3}}$$

D

$$\frac{2}{9}$$

View Explanation Hide Explanation

Solution

$$\sin^6\theta+\cos^6\theta=\frac{1}{3}$$

$$\left(\sin^2\theta\right)^3+\left(\cos^2\theta\right)^3=\frac{1}{3}$$

$$\left(\sin^2\theta+\cos^2\theta\right)\left(\sin^4\theta-\sin^2\theta\ \cos^2\theta+\cos^4\theta\ \right)=\frac{1}{3}$$

$$\left(1\right)\left(\sin^4\theta+\cos^4\theta+2\sin^2\theta\ \cos^2\theta-3\sin^2\theta\ \cos^2\theta\ \right)=\frac{1}{3}$$

$$\left(\sin^2\theta+\cos^2\theta\right)^2-3\sin^2\theta\ \cos^2\theta\ =\frac{1}{3}$$

$$1-3\sin^2\theta\ \cos^2\theta\ =\frac{1}{3}$$

$$3\sin^2\theta\ \cos^2\theta\ =\frac{2}{3}$$

$$\sin^2\theta\ \cos^2\theta\ =\frac{2}{9}$$

$$\sin\theta\ \cos\theta=\frac{\sqrt{2}}{3}$$

Hence, the correct answer is Option A

Question 56 Show answer

The value of $$\sin^2 60^\circ \cos^2 45^\circ + 2 \tan^2 60^\circ - \cosec^2 30^\circ$$ is equal to:

A

$$\frac{17}{24}$$

B

$$\frac{19}{8}$$

C

$$-\frac{17}{24}$$

D

$$-\frac{19}{8}$$

View Explanation Hide Explanation

Solution

$$\sin^260^{\circ}\cos^245^{\circ}+2\tan^260^{\circ}-\operatorname{cosec}^230^{\circ}=\left(\frac{\sqrt{3}}{2}\right)^2.\left(\frac{1}{\sqrt{2}}\right)^2+2\left(\sqrt{3}\right)^2-\left(2\right)^2$$

$$=\frac{3}{4}.\frac{1}{2}+2\left(3\right)-4$$

$$=\frac{3}{8}+6-4$$

$$=\frac{3}{8}+2$$

$$=\frac{3+16}{8}$$

$$=\frac{19}{8}$$

Hence, the correct answer is Option B

Question 57 Show answer

If $$4 \sin^2 (2x - 10)^\circ = 3, 0 \leq (2x - 10) \leq 90$$, then find the value of $$\frac{\sin^4(x - 5)^\circ + \cos^4 (x - 5)^\circ}{1 - 2 \sin^2 (3x - 15)^\circ \cos^2(3x - 15)^\circ}$$

A

$$\frac{5}{8}$$

B

$$-1$$

C

$$1$$

D

$$-\frac{5}{8}$$

Question 58 Show answer

Find the value of $$\sin^4 30^\circ + \cos^4 30^\circ - \sin 25^\circ \cos 65^\circ - \sin 65^\circ \cos 25^\circ$$

A

$$-\frac{3}{8}$$

B

0

C

$$\frac{5}{8}$$

D

$$\frac{13}{8}$$

Question 59 Show answer

If $$0^\circ < \theta < 90^\circ, \sqrt{\frac{\sec^2 \theta + \cosec^2 \theta}{\tan^2 \theta - \sin^2 \theta}}$$ is equal to:

A

$$\sec^3 \theta$$

B

$$\cosec^3 \theta$$

C

$$\sin^2 \theta$$

D

$$\sec^2 \theta$$

View Explanation Hide Explanation

Solution

$$\sqrt{\frac{\sec^2\theta+\operatorname{cosec}^2\theta}{\tan^2\theta-\sin^2\theta}}=\sqrt{\frac{\frac{1}{\cos^2\theta}+\frac{1}{\sin^2\theta}}{\frac{\sin^2\theta\ }{\cos^2\theta}-\sin^2\theta}}$$

$$=\sqrt{\frac{\frac{\sin^2\theta+\cos^2\theta\ }{\cos^2\theta\sin^2\theta}}{\frac{\sin^2\theta-\sin^2\theta\cos^2\theta\ }{\cos^2\theta}}}$$

$$=\sqrt{\frac{1\ }{\cos^2\theta\sin^2\theta}\times\frac{\cos^2\theta\ }{\sin^2\theta\left(1-\cos^2\theta\right)\ }}$$

$$=\sqrt{\frac{1\ }{\sin^2\theta}\times\frac{1}{\sin^2\theta\left(\sin^2\theta\right)\ }}$$

$$=\sqrt{\frac{1\ }{\sin^6\theta}\ }$$

$$=\sqrt{\operatorname{cosec}^6\theta\ }$$

$$=\operatorname{cosec}^3\theta\ $$

Hence, the correct answer is Option B

Question 60 Show answer

If $$3 \sec \theta + 4 \cos \theta - 4\sqrt{3} = 0$$ where $$\theta$$ is an acute angle then the value of $$\theta$$ is:

A

$$20^\circ$$

B

$$30^\circ$$

C

$$60^\circ$$

D

$$45^\circ$$

View Explanation Hide Explanation

Solution

$$3\sec\theta+4\cos\theta-4\sqrt{3}=0$$

$$\frac{3}{\cos\theta\ }+4\cos\theta-4\sqrt{3}=0$$

$$4\cos^2\theta-4\sqrt{3}\cos\theta\ +3=0$$

$$4\cos^2\theta-2\sqrt{3}\cos\theta\ -2\sqrt{3}\cos\theta+3=0$$

$$2\cos\theta\ \left(2\cos\theta-\sqrt{3}\right)\ -\sqrt{3}\left(2\cos\theta-\sqrt{3}\right)=0$$

$$\ \left(2\cos\theta-\sqrt{3}\right)\left(2\cos\theta-\sqrt{3}\right)=0$$

$$\ \left(2\cos\theta-\sqrt{3}\right)^2=0$$

$$\ 2\cos\theta-\sqrt{3}=0$$

$$\cos\theta=\frac{\sqrt{3}}{2}$$

$$\theta=30^{\circ\ }$$

Hence, the correct answer is Option B

Question 61 Show answer

If $$\sin \alpha + \sin \beta = \cos \alpha + \cos \beta = 1$$, then $$\sin \alpha + \cos \alpha =$$?

A

-1

B

0

C

1

D

2

View Explanation Hide Explanation

Solution

$$\sin\alpha+\sin\beta=1$$

$$\sin^2\alpha+\sin^2\beta+2\sin\alpha\ \sin\beta\ =1$$......(1)

$$\cos\alpha+\cos\beta=1$$

$$\cos^2\alpha+\cos^2\beta+2\cos\alpha\ \cos\beta\ =1$$......(2)

Adding (1) and (2),

$$\left(\sin^2\alpha\ +\cos^2\alpha\right)+\left(\sin^2\beta\ +\cos^2\beta\right)+2\sin\alpha\ \sin\beta\ +2\cos\alpha\ \cos\beta\ =1+1$$

$$1+1+2\sin\alpha\ \sin\beta\ +2\cos\alpha\ \cos\beta\ =2$$

$$2\left[\cos\alpha\ \cos\beta+\sin\alpha\ \sin\beta\right]=0$$

$$\cos\left(\beta-\alpha\right)=0$$

$$\beta-\alpha=90^{\circ\ }$$

$$\beta\ =90^{\circ\ }+\alpha\ $$

$$\sin\alpha+\sin\beta=1$$

$$\sin\alpha+\sin\left(90^{\circ}-\alpha\ \right)=1$$

$$\sin\alpha+\cos\alpha=1$$

Hence, the correct answer is Option C

Question 62 Show answer

In $$\triangle PQR, \angle Q = 90^\circ$$. If $$\tan R = \frac{1}{3}$$, then what is the value of $$\frac{\sec P(\cos R + \sin P)}{\cosec R(\sin R - \cosec P)}$$?

A

$$\frac{18}{7}$$

B

$$-\frac{18}{7}$$

C

$$-\frac{2}{7}$$

D

$$\frac{2}{7}$$

Question 63 Show answer

If $$2 \sin (3x - 15)^\circ = 1, 0^\circ < (3x - 15) < 90^\circ$$, then find the value of $$\cos^2 (2x + 15)^\circ + \cot^2 (x + 15)^\circ$$.

A

$$\frac{7}{2}$$

B

$$1$$

C

$$-\frac{7}{2}$$

D

$$\frac{5}{2}$$

Question 64 Show answer

If x is a real quantity, whatis the minimum value of $$(25 \cos^2 x + 9 \sec^2 x)$$

A

30

B

20

C

15

D

40

Question 65 Show answer

The value of $$\frac{\sqrt{2}\tan(60^\circ - \theta) \tan(30^\circ + \theta)}{\sin^2 (45^\circ + \theta) + \sin^2 (45^\circ - \theta)}$$ is:

A

$$\sqrt{2}$$

B

2

C

1

D

$$\frac{1}{\sqrt{2}}$$

Question 66 Show answer

The value of $$\frac{\tan(45^\circ - \alpha)}{\cot(45^\circ + \alpha)} - \frac{(\cos 19^\circ + \sin 71^\circ)(\sec 19^\circ + \cosec 71^\circ)}{\tan 12^\circ \tan 24^\circ \tan 66^\circ \tan 78^\circ}$$ is:

A

-3

B

0

C

-2

D

2

Question 67 Show answer

The value of $$\sec^4 \theta (1 - \sin^4 \theta) - 2 \tan^2 \theta$$ is:

A

$$\frac{1}{2}$$

B

1

C

-1

D

0

View Explanation Hide Explanation

Solution

$$\sec^4\theta(1-\sin^4\theta)-2\tan^2\theta=\sec^4\theta\ -\sec^4\theta\ \sin^4\theta\ -2\tan^2\theta\ $$

$$=\sec^4\theta\ -\frac{\sin^4\theta}{\cos^4\theta\ }\ -2\tan^2\theta\ $$

$$=\sec^4\theta\ -\tan^4\theta-2\tan^2\theta\ $$

$$=\left(\sec^2\theta\ +\tan^2\theta\ \right)\left(\sec^2\theta\ -\tan^2\theta\ \right)-2\tan^2\theta\ $$

$$=\left(\sec^2\theta\ +\tan^2\theta\ \right)\left(1\right)-2\tan^2\theta\ $$

$$=\sec^2\theta\ +\tan^2\theta\ -2\tan^2\theta\ $$

$$=\sec^2\theta\ -\tan^2\theta\ $$

$$=1$$

Hence, the correct answer is Option B

Question 68 Show answer

Find the value of $$\sin^2 60^\circ + \cos^2 30^\circ - \sin^2 45^\circ - 3 \sin^2 90^\circ$$.

A

$$\frac{1}{3}$$

B

$$-1\frac{3}{4}$$

C

$$-2\frac{1}{2}$$

D

$$-2$$

View Explanation Hide Explanation

Solution

$$\sin^260^{\circ}+\cos^230^{\circ}-\sin^245^{\circ}-3\sin^290^{\circ}=\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2-\left(\frac{1}{\sqrt{2}}\right)^2-3\left(1\right)^2$$

$$=\frac{3}{4}+\frac{3}{4}-\frac{1}{2}-3$$

$$=\frac{3+3-2-12}{4}$$

$$=\frac{-8}{4}$$

$$=-2$$

Hence, the correct answer is Option D

Question 69 Show answer

If $$\cos(A - B) = \frac{\sqrt{3}}{2}$$ and $$\cot(A + B) = \frac{1}{\sqrt{3}}$$ Where A - B and A + B are acute angles, then (2A - 3B) is equal to:

A

$$45^\circ$$

B

$$30^\circ$$

C

$$15^\circ$$

D

$$60^\circ$$

Question 70 Show answer

$$(\operatorname{cosec}A-\cot A)(1+\cos A)=?$$

A

$$\cos A$$

B

$$\sin A$$

C

$$\cot A$$

D

$$\cosec A$$

View Explanation Hide Explanation

Solution

$$(\operatorname{cosec}A-\cot A)(1+\cos A)=\left(\frac{1}{\sin A}-\frac{\cos A}{\sin A}\right)\left(1+\cos A\right)$$

= $$\left(\frac{1-\cos A}{\sin A}\right)\left(1+\cos A\right)$$

= $$\left(\frac{1-\cos^2A}{\sin A}\right)$$

= $$\frac{\sin^2A}{\sin A}$$

= $$\sin A$$

Hence, the correct answer is Option B

Question 71 Show answer

$$(\sec \theta + \tan \theta)^2 + \frac{1 + \cosec \theta}{1 - \cosec \theta}, 0^\circ < \theta < 90^\circ$$ is:

A

0

B

-2

C

1

D

2

Question 72 Show answer

Simplify $$\sec^2 \alpha \left(1 + \frac{1}{\cosec \alpha}\right)\left(1 - \frac{1}{\cosec \alpha}\right)$$.

A

$$\tan^4 \alpha$$

B

$$\sin^2 \alpha$$

C

1

D

-1

View Explanation Hide Explanation

Solution

$$\sec^2\alpha\left(1+\frac{1}{\operatorname{cosec}\alpha}\right)\left(1-\frac{1}{\operatorname{cosec}\alpha}\right)=\frac{1}{\cos^2\alpha}\left(1+\sin\alpha\right)\left(1-\sin\alpha\right)$$

$$=\frac{1}{\cos^2\alpha}\left(1-\sin^2\alpha\right)$$

$$=\frac{1}{\cos^2\alpha}\left(\cos^2\alpha\right)$$

$$=1$$

Hence, the correct answer is Option C

Question 73 Show answer

If $$2 \cos^2 \theta = 3 \sin \theta, 0^\circ < \theta < 90^\circ$$, then the value of $$(sec^2 \theta - \tan^2 \theta + \cos^2 \theta)$$ is:

A

$$\frac{9}{4}$$

B

$$\frac{5}{4}$$

C

$$\frac{3}{4}$$

D

$$\frac{7}{4}$$

Question 74 Show answer

The value of $$\frac{\tan^2 30^\circ + \sin^2 90^\circ + \cot^2 60^\circ + \sin^2 30^\circ \cos^2 45^\circ}{\sin 60^\circ \cos 30^\circ - \cos 60^\circ \sin 30^\circ}$$ is:

A

$$\frac{47}{12}$$

B

$$\frac{37}{12}$$

C

$$\frac{43}{12}$$

D

$$\frac{25}{12}$$

Question 1 Show answer

As observed from the top of a lighthouse, 45 m high above the sea-level, the angle of depression of a ship, sailing directly towardsit, changes from $$30^\circ$$ to $$45^\circ$$. The distance travelled by the ship during the period of observation is: (Your answer should be correct to one decimalplace.)

A

32.9 m

B

33.4 m

C

36.9 m

D

24.8 m

View Explanation Hide Explanation

Solution

Let AB be the light house of height 45 m. Let CD = $$x$$ m

In right $$\triangle$$ ABC,

=> $$tan(\angle ACB)=\frac{AB}{BC}$$

=> $$tan(45^\circ)=1=\frac{45}{BC}$$

=> $$BC=45$$ m

In right $$\triangle$$ ABD,

=> $$tan(\angle ADB)=\frac{AB}{BD}$$

=> $$tan(30^\circ)=\frac{1}{\sqrt3}=\frac{45}{45+x}$$

=> $$x+45=45\sqrt3$$

=> $$x=45(\sqrt3-1)$$

=> $$x=45\times0.732=32.9$$ m

=> Ans - (A)

Question 2 Show answer

A 22 m long ladder (whose foot is on the ground) leans against a wall making an angle of $$60^\circ$$ with the ground. What is the height (in m) of the point where the ladder touches the wall from the ground?

A

$$\frac{22\sqrt{2}}{3}$$

B

$$11\sqrt{2}$$

C

11

D

$$11\sqrt{3}$$

View Explanation Hide Explanation

Solution

Let AC be the ladder of height 22 m and $$\angle$$ ACB = $$60^\circ$$

In right $$\triangle$$ ABC,

=> $$sin(\angle ACB)=\frac{AB}{AC}$$

=> $$sin(60^\circ)=\frac{\sqrt3}{2}=\frac{h}{22}$$

=> $$h=11\sqrt3$$ m

=> Ans - (D)

Question 3 Show answer

The angles of elevation of a pole from two points which are 75 m and 48 m away from its base are $$\alpha$$ and $$\beta$$, respectively. If $$\alpha$$ and $$\beta$$ are complementary, then the height of the tower is:

A

54.5 m

B

61.5 m

C

60 m

D

50 m

View Explanation Hide Explanation

Solution

$$\angle \alpha +\angle \beta = 90\degree$$ (complementary angle)

OA = 48 m

OB = 75 m

$$tan\beta = \frac{OP}{OA}$$

$$tan\beta = \frac{OP}{48}$$

$$tan\alpha = \frac{OP}{75}$$

$$tan(90 - \beta) = \frac{OP}{75}$$

$$cot\beta = \frac{OP}{75}$$

$$tan\beta = \frac{1}{cot\beta}$$

$$\frac{OP}{48} = \frac{75}{OP}$$

OP = $$\sqrt{75 \times 48} = \sqrt{3600}$$ = 60 m

The height of tower is 60 m.

Question 4 Show answer

If $$3 + \cos^2 \theta = 3(\cot^2 \theta + \sin^2 \theta), 0^\circ < \theta < 90^\circ$$, then what is the value of $$(\cos \theta + 2 \sin \theta)$$?

A

$$3\sqrt{2}$$

B

$$\frac{\sqrt3 + 2}{2}$$

C

$$\frac{2\sqrt{3} + 1}{2}$$

D

$$\frac{3\sqrt{3} + 1}{2}$$

View Explanation Hide Explanation

Solution

Given : $$3 + \cos^2 \theta = 3(\cot^2 \theta + \sin^2 \theta), 0^\circ < \theta < 90^\circ$$

=> $$3+(1-sin^2\theta)=3(cosec^2\theta-1)+3sin^2\theta$$

=> $$4-4sin^2\theta=\frac{3}{sin^2\theta}-3$$

=> $$7sin^2\theta-4sin^4\theta=3$$

Let $$sin^2\theta=x$$

=> $$4x^2-7x+3=0$$

=> $$(x-1)(4x-3)=0$$

=> $$x=1,\frac{3}{4}$$

=> $$sin\theta=1,\frac{\sqrt3}{2}$$

$$\because \theta<90^\circ$$, => $$\theta=60^\circ$$

$$\therefore$$ $$(\cos \theta + 2 \sin \theta)$$

= $$cos(60^\circ)+2sin(60^\circ)$$

= $$\frac{1}{2}+(2\times\frac{\sqrt3}{2})$$

= $$\frac{2\sqrt3+1}{2}$$

=> Ans - (C)

Question 5 Show answer

A kite is flying at a height of 123 m. The thread attached to it is assumed to bestretched straight and makes an angle of $$60^\circ$$ with the level ground. The length ofthe string is (nearest to a whole number):

A

140 m

B

139 m

C

142 m

D

138 m

View Explanation Hide Explanation

Solution

Let AC be the string of kite = $$x$$ m

In right $$\triangle$$ ABC,

=> $$sin(\angle ACB)=\frac{AB}{AC}$$

=> $$sin(60^\circ)=\frac{\sqrt3}{2}=\frac{123}{x}$$

=> $$x=\frac{246}{\sqrt3}$$

=> $$x=82\sqrt3\approx142$$ m

=> Ans - (C)

Question 6 Show answer

Aclock tower stands at the crossing of two roads whichpoint in the north-south and the east-westdirections. P, Q, R and S are points on the roads due north, east, south and west respectively, where the angles of elevation of the top of the tower are respectively, $$\alpha, \beta, \gamma$$ and $$\delta$$, Then $$\left(\frac{PQ}{RS}\right)^2$$ is equal to:

A

$$\frac{\tan^2 \alpha + \tan^2 \delta}{\tan^2 \beta + \tan^2 \gamma}$$

B

$$\frac{\tan^2 \alpha + \tan^2 \beta}{\tan^2 \gamma + \tan^2 \delta}$$

C

$$\frac{\cot^2 \alpha + \cot^2 \beta}{\cot^2 \gamma + \cot^2 \delta}$$

D

$$\frac{\cot^2 \alpha + \cot^2 \delta}{\cot^2 \beta + \cot^2 \gamma}$$

View Explanation Hide Explanation

Solution

From the give question, we draw the diagram is given below

Now by the 3D view 

Let the height of tower = h

$$ \frac{h}{x} = \tan \alpha $$ ......(1)  ( alpha angle of elevation of tower with point P)

Similarly,  $$\frac{h}{y} = \tan \beta $$ ..... (2)

$$\frac {h}{z} = \tan \gamma $$ .......(3)

$$\frac{h}{a} = \tan \rho$$ -- (4)

From Equestion  (1)(2)(3)(4) 

$$x = \frac{h} {\tan \alpha} , a= \frac{h} {\tan \rho} , y= \frac{h} {\tan \beta} , z= \frac{h}{\tan \gamma} $$

Now PQ = $$ \sqrt {x^2 +y^2} $$

           = $$\sqrt{ \frac {h^2}{\tan^2 \alpha} + \frac{h^2}{ \tan^2 \beta} }$$  ...... Equestion (5) 

RS = $$\sqrt { z^2 + a^2} $$ 

   = $$\sqrt{\frac{h^2} {\tan^2 \gamma} + \frac{h^2} {\tan^2 \rho } }$$ ......... equestion (6)

Equestion (5) divided by Equestion (6)

$$ (\frac{PQ}{RS} )^2= \frac{ \cot^2 \alpha + \cot^2 \beta} { \cot^2 \gamma + \cot^2 \rho } $$

therefore Option (C) Ans 

Question 7 Show answer

If $$\cos^2 \theta - \sin^2 \theta = \tan^2 \phi$$ then which of the following is true?

A

$$\cos \theta \cos \phi =1$$

B

$$\cos^2 \phi - \sin^2 \phi = \tan^2 \theta$$

C

$$\cos^2 \phi - \sin^2 \phi = \cot^2 \theta$$

D

$$\cos \theta \cos \phi = \sqrt{2}$$

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Solution

Expression : $$\cos^2 \theta - \sin^2 \theta = \tan^2 \phi$$

=> $$cos^2\theta-sin^2\theta=\frac{sin^2\phi}{cos^2\phi}$$

=> $$\frac{cos^2\theta-sin^2\theta}{cos^2\theta+sin^2\theta}=\frac{sin^2\phi}{cos^2\phi}$$

By componendo and dividendo

=> $$\frac{-sin^2\theta}{cos^2\theta}=\frac{sin^2\phi-cos^2\phi}{sin^2\phi+cos^2\phi}$$

=> $$cos^2\phi-sin^2\phi=tan^2\theta$$

=> Ans - (B)

Question 8 Show answer

The angle of elevation ofthe top of a tree from a point on the ground which is 300 m away from thetree is $$30^\circ$$. When the tree grew up, its angle ofelevation ofthe top of it became $$60^\circ$$ from the same point. How much did the tree grow? (nearest to an integer)

A

342 m

B

364 m

C

384 m

D

346 m

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Solution

we draw the diagram is given below

Ab is the height that tree grew up 

Let AB= h 

In $$\triangle BCD $$

$$\frac{BC} {CD} = \tan \30^\circ $$

BC =$$ 300 \times \frac{1}{\sqrt {3}} $$ 

BC = $$100 \sqrt {3}$$ m  

In $$ \triangle ACD $$

$$ \frac{AC} {CD} = \tan 60^\circ $$

$$ \Rightarrow \frac{AB+BC} {300} = \sqrt {3} $$

$$ AB + 100 \sqrt {3} = 300 \sqrt {3} $$

$$ AB = 300 \sqrt{3} - 100 \sqrt{3} $$

$$ AB = 200 \sqrt{3} $$ 

$$ AB = 200 \times 1.73 $$

$$ AB = 346 $$ 

therefore the tree grow = 346 m Ans 

Question 9 Show answer

If $$\tan (11 \theta) = \cot (7 \theta)$$, then what is the value of $$\sin^2 (6 \theta) + \sec^2 (9 \theta) + \cosec^2 (12 \theta)$$?

A

$$\frac{35}{12}$$

B

$$\frac{23}{6}$$

C

$$\frac{31}{12}$$

D

$$\frac{43}{12}$$

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Solution

Given : $$\tan (11 \theta) = \cot (7 \theta)$$

=> $$\tan (11 \theta) = \tan (90^\circ-7 \theta)$$

=> $$11\theta=90^\circ-7\theta$$

=> $$11\theta+7\theta=18\theta=90^\circ$$

=> $$\theta=5^\circ$$

To find : $$\sin^2 (6 \theta) + \sec^2 (9 \theta) + \cosec^2 (12 \theta)$$

= $$sin^2(30^\circ)+sec^2(45^\circ)+cosec^2(60^\circ)$$

= $$\frac{1}{4}+2+\frac{4}{3}$$

= $$\frac{3+24+16}{12}=\frac{43}{12}$$

=> Ans - (D)

Question 10 Show answer

If $$\cosec \theta = 1.25,$$ then $$\frac{4 \tan \theta - 5 \cos \theta + 1}{\sec \theta + 4 \cot \theta -1} = ?$$

A

$$\frac{1}{2}$$

B

$$\frac{9}{10}$$

C

$$2$$

D

$$\frac{10}{11}$$

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Solution

Given that $$\cosec \theta = 1.25  = \dfrac {5}{4}$$

so $$\sin \theta = \dfrac{4}{5} $$

then $$\cos \theta = \dfrac {3}{5} $$, $$\tan \theta = \dfrac {4}{3}$$, $$\cot \theta = \dfrac{3}{4}$$

Hence $$ \dfrac {4 \tan \theta - 5 \cos \theta + 1}{ \sec \theta + 4 \cot \theta -1}$$

$$\Rightarrow \dfrac {4\times \dfrac{4}{3} - 5 \times \dfrac{3}{5} +1} {\dfrac{5}{3} + 4\times \dfrac{3}{4} -1}$$

$$\Rightarrow \dfrac{ \dfrac{16}{3} -2} {\dfrac {5} {3} +2} $$

$$\Rightarrow \dfrac{16-6}{5+6}$$

$$\Rightarrow \dfrac{10}{11} $$ Ans 

Question 11 Show answer

A ladder leaning against a window of a house makes an angle of 60° with the ground. Ifthe distance of the foot of the ladder from the wall is 4.2 m, then the height of the point, where the ladder touches the window from the ground is Closest to:

A

7. 3 m

B

6.8 m

C

7.8 m

D

7 m

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Solution

As per the given question,

The ladder is making $$60^\circ$$ with the ground

and the foot of the ladder is (x)=4.2m from the wall.
Height of the wall till window (h)=?

We know that $$\tan\theta=\dfrac{h}{x}$$

$$\tan 60^\circ=\dfrac{h}{4.2}$$

$$x=\sqrt{3}\times 4.2=1.732\times 4.2\simeq=7.3 m$$

Question 12 Show answer

If $$\frac{\cos \theta}{1 - \sin \theta} + \frac{\cos \theta}{1 + \sin \theta} = 4, 0^\circ < \theta < 90^\circ$$ then what is the value of $$(\sec \theta + \cosec \theta + \cot \theta)$$?

A

$$1 + 2\sqrt{3}$$

B

$$\frac{1 + 2\sqrt{3}}{3}$$

C

$$\frac{2 + \sqrt{3}}{3}$$

D

$$2 + \sqrt{3}$$

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Solution

$$\frac{\cos \theta}{1 - \sin \theta} + \frac{\cos \theta}{1 + \sin \theta} = 4$$

$$\frac{cos\theta + sin\theta cos\theta+cos\theta-sin\theta cos\theta}{sin^2\theta - 1}$$ = 4

$$\frac{2cos\theta}{cos^2\theta}$$ = 4

$$cos\theta = \frac{1}{2}$$

$$\theta = 60\degree$$

$$(\sec \theta + \cosec \theta + \cot \theta)$$

= $$(\sec60\degree+ \cosec60\degree + \cot60\degree)$$

= 2 + $$\frac{2}{\sqrt3} + \frac{1}{\sqrt3} = \frac{2\sqrt{3} + 2 + 1}{3} = 2 + \sqrt{3}$$

Question 13 Show answer

If $$2\sin \theta + 15 \cos^2 \theta = 7, 0^\circ < \theta < 90^\circ$$ then what is the value of $$\frac{3 - \tan \theta}{2 + \tan \theta}$$?

A

$$\frac{1}{4}$$

B

$$\frac{5}{8}$$

C

$$\frac{1}{2}$$

D

$$\frac{3}{4}$$

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Solution

Given that  $$2 \sin \theta + 15 cos^2  \theta =7 $$

$$\Rightarrow 2\sin \theta  + 15 ( 1- \sin^2 \theta ) - 7 = 0 $$

$$\Rightarrow 15 \sin^2 \theta - 2 \sin \theta - 8 = 0 $$

$$ \Rightarrow 15 \sin^2 \theta + 10 \sin \theta - 12 \sin \theta - 8 = 0 $$ 

$$\Rightarrow  (3 \sin \theta + 2) (5\sin \theta - 4) = 0 $$ 

$$\Rightarrow \sin \theta = \dfrac{4}{5} $$ ,$$ \sin \theta = - \dfrac{2}{3} $$

then valid value $$\sin \theta = \dfrac {4}{5}$$ 

so  $$\cos \theta = \dfrac {3}{4} $$, $$\tan \theta = \dfrac {4}{3}$$ 

hence $$\dfrac {3 - \tan \theta} {2 + \tan \theta }$$

$$\Rightarrow \dfrac {3 - \dfrac{4}{3}} { 2 + \dfrac {4} {3}}$$ 

$$\Rightarrow \dfrac {9- 4}{6+4}$$

$$\Rightarrow \dfrac {5}{10} $$

$$\Rightarrow \dfrac {1}{2} $$ Ans 

Question 14 Show answer

If $$0 \leq \theta \leq 90^\circ$$, and $$(2\theta + 50^\circ) = \cos (4\theta + 16^\circ)$$, then what is the value of $$\theta$$ (in degrees)?

A

$$8^\circ$$

B

$$10^\circ$$

C

$$12^\circ$$

D

$$4^\circ$$

Question 15 Show answer

A pole of length 7 m is fixed vertically on the top of a tower. The angle of elevation of the top of the pole observed from a point on the ground is $$60^\circ$$ and the angle of depression of the same point on the ground from the top of the tower is $$45^\circ$$. The height (in m) of the tower is:

A

$$7(2\sqrt{3} - 1)$$

B

$$\frac{7}{2}(\sqrt{3} + 2)$$

C

$$7\sqrt{3}$$

D

$$\frac{7}{2}(\sqrt{3} + 1)$$

View Explanation Hide Explanation

Solution

As per the given question,

Length of the pole is DC=7m
The angle of elevation of the top of the pole to ground$$=60^\circ$$

The angle of elevation of the top of the tower $$=45^\circ$$

Now, $$\tan 60^\circ=\dfrac{BC}{AB}$$

$$\Rightarrow \sqrt{3}=\dfrac{BC}{AB}$$

$$\Rightarrow AB=\dfrac{BC}{\sqrt{3}}------(i)$$

Now, In the $$\triangle ABD$$

$$\Rightarrow \tan 45^\circ=\dfrac{BD}{AB}$$

$$\Rightarrow 1=\dfrac{BD}{AB}$$

$$\Rightarrow AB=BD------(ii)$$

From (i) and (ii)

$$\Rightarrow BD=\dfrac{DC+BD}{\sqrt{3}}$$

$$\Rightarrow BD=\dfrac{7+BD}{\sqrt{3}}$$

$$\Rightarrow BD\sqrt{3}=7+BD$$

$$\Rightarrow BD(\sqrt{3}-1)=7$$

$$\Rightarrow BD=\dfrac{7}{\sqrt{3}-1}$$

$$\Rightarrow BD=\dfrac{7(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$$

$$\Rightarrow BD=\dfrac{7(\sqrt{3}+1)}{3-1}$$

$$\Rightarrow BD=\dfrac{7(\sqrt{3}+1)}{2}$$

Question 16 Show answer

The angle of elevation of a ladder leaning against a wall is $$60^\circ$$ and the foot of the ladder is 6.5 m away from the wall. The length of the ladder is:

A

$$6.5 \sqrt{3}$$ m

B

$$13 \sqrt{3}$$ m

C

12 m

D

13 m

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Solution

from the above question we draw the diagram 

Let Lenght ladder = AC

then from the  $$ \triangle ABC $$,

    $$ \cos 60^\circ =  \frac {AB} {AC} $$

     $$\Rightarrow \frac{1}{2} = \frac {6.5} {AC}$$

      $$\Rightarrow AC = 13  m $$ 

therefore The length of the ladder = 13 Ans 

Question 17 Show answer

If $$7 \sin^2 \theta + 3 \cos^2 \theta = 4, 0^\circ < \theta < 90^\circ$$, then the value of $$(\tan^2 2 \theta + \cosec^2 2 \theta)$$ is:

A

7

B

$$\frac{15}{4}$$

C

$$\frac{13}{3}$$

D

$$\frac{13}{4}$$

View Explanation Hide Explanation

Solution

Expression : $$7 \sin^2 \theta + 3 \cos^2 \theta = 4, 0^\circ < \theta < 90^\circ$$

=> $$7sin^2\theta+3(1-sin^2\theta)=4$$

=> $$4sin^2\theta=1$$

=> $$sin\theta=\sqrt{\frac{1}{4}}$$

=> $$\theta=sin^{-1}(\frac{1}{2})=30^\circ$$

To find : $$(\tan^2 2 \theta + \cosec^2 2 \theta)$$

= $$tan^2(60^\circ)+cosec^2(60^\circ)$$

= $$3+\frac{4}{3}=\frac{13}{3}$$

=> Ans - (C)

Question 18 Show answer

If $$\frac{1}{\cosec \theta + 1} + \frac{1}{\cosec \theta -1} = 2 \sec \theta, 0^\circ < \theta < 90^\circ$$, then the value of $$\frac{\tan \theta + 2 \sec \theta}{\cosec \theta}$$ is:

A

$$\frac{2+\sqrt{3}}{2}$$

B

$$\frac{4+\sqrt{2}}{2}$$

C

$$\frac{2+\sqrt{2}}{2}$$

D

$$\frac{4+\sqrt{3}}{2}$$

View Explanation Hide Explanation

Solution

Given that $$\dfrac{1}{\cosec\theta +1} + \dfrac {1}{\cosec \theta -1} = 2 \sec \theta $$

$$\Rightarrow \dfrac {\cosec \theta - 1 + \cosec \theta +1}{ \cosec^2 \theta -1} = 2\sec \theta $$

$$\Rightarrow \dfrac{2 \cosec \theta}{\cot^2 \theta} = 2 \sec \theta $$

$$\Rightarrow \dfrac {1}{\sin \theta} \times \dfrac {\sin^2 \theta}{\cos^2 \theta} = \dfrac {1}{\cos \theta} $$

$$\Rightarrow \dfrac{\sin \theta}{\cos \theta} = 1 $$

$$\Rightarrow  \tan \theta =1$$

$$\Rightarrow \theta = \dfrac{\pi}{4}$$

then value $$\dfrac {\tan \theta + 2 \sec \theta}{\cosec \theta}$$

$$\Rightarrow \dfrac {\tan \dfrac{\pi}{4} + 2 \sec \dfrac {\pi}{4}}{\cosec \dfrac{\pi}{4}}$$  (put the value $$\theta $$)

$$\Rightarrow \dfrac{1 + 2\sqrt{2}} {\sqrt {2}}$$ multiply by $$ \sqrt {2}$$

$$\Rightarrow \dfrac {4 + \sqrt {2}}{2} $$ Ans

Question 19 Show answer

An observer who is 1.62 m tall is 45 m away from a pole. The angle of elevation of the top of the pole from his eyes is $$30^\circ$$. The height (in m) ofthe pole is closest to:

A

26.2

B

27.6

C

25.8

D

26.8

View Explanation Hide Explanation

Solution

Given that observer 1.62 m away from 45 m  angle $$30^\circ$$

AB= 45 m

then in the diagram

from the above figure  $$\tan 30^\circ = \frac{h-1.62} {45} $$

$$\Rightarrow h-1.62 = 45 \times \frac{1} {\sqrt{3}} $$

$$\Rightarrow h= 45 \times \frac{1} {\sqrt{3}} + 1.62 $$

$$\Rightarrow h= 45 \times 0.577 +1.62 $$

$$\Rightarrow h= 27.58 $$ m 

$$\Rightarrow h= 27.6  $$ m Ans 

Question 20 Show answer

If $$21 \tan \theta = 20$$, then $$(1 + \sin \theta + \cos \theta) : (1 - \sin \theta + \cos \theta) = ?$$

A

5 : 2

B

3 : 1

C

7 : 3

D

2 : 1

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Solution

Given that $$21 \tan \theta = 20$$

$$ \tan \theta = \dfrac{20}{21}$$

then $$(1 + \sin \theta + \cos \theta) : (1 - \sin \theta + \cos \theta)$$

$$\dfrac{(1 + \sin \theta + \cos \theta)}{ (1 - \sin \theta + \cos \theta)}$$

dividing   $$\cos \theta $$

$$\dfrac {(\sec \theta + \tan \theta + 1)}{(\sec \theta - \tan \theta + 1)} $$

we know that $$\sec^2 \theta - \tan^2 \theta = 1 $$

then $$\Rightarrow \dfrac{(\sec \theta +\tan \theta) (\sec^2 \theta - \tan^2 \theta)} {(\sec \theta - \tan \theta + 1)}$$

$$\Rightarrow (\sec \theta + \tan \theta) $$

Now $$ \tan \theta = \dfrac{20}{21} $$

we know that $$ 1 + \tan^2 \theta = \sec^2 \theta $$

$$\Rightarrow 1 + \dfrac{400}{441} = \sec^2 \theta $$ 

$$\Rightarrow \sec \theta = \dfrac{29}{21}$$

hence $$\sec \theta + \tan \theta $$

$$\Rightarrow \dfrac{29}{21}+ \dfrac{20}{21}$$ (put the value)

$$\Rightarrow \dfrac{49}{21}$$

$$\Rightarrow \dfrac{7}{3} $$ Ans

Question 21 Show answer

If $$3(\cot^2 \theta - \cos^2 \theta) = 1 - \sin^2 \theta, 0^\circ < \theta < 90^\circ$$, then the $$\theta$$ equal to

A

$$30^\circ$$

B

$$60^\circ$$

C

$$45^\circ$$

D

$$15^\circ$$

View Explanation Hide Explanation

Solution

$$3(\cot^2 \theta - \cos^2 \theta) = 1 - \sin^2 \theta, 0^\circ < \theta < 90^\circ$$

$$\Rightarrow 3(\dfrac{\cos^2\theta}{\sin^2\theta} - \cos^2 \theta) = 1 - \sin^2 \theta$$

$$\Rightarrow 3\cos^2\theta(\dfrac{1-\sin^2\theta}{\sin^2\theta} ) = 1 - \sin^2 \theta$$

$$\Rightarrow \tan^2\theta=3$$

$$\Rightarrow \tan\theta=\sqrt{3}=\tan 60^\circ$$

$$\theta=60^\circ$$

Question 22 Show answer

If $$(\cos \theta + \sin \theta) : (\cos \theta - \sin \theta) = (\sqrt{3} + 1) : (\sqrt{3} - 1), 0^\circ < \theta < 90^\circ$$, then what is the value of $$\sec \theta$$?

A

1

B

2

C

$$\sqrt{2}$$

D

$$\frac{2\sqrt3}{3}$$

View Explanation Hide Explanation

Solution

Given that $$(\cos \theta + \sin \theta) : (\cos \theta - \sin \theta) = (\sqrt{3} + 1) : (\sqrt{3} - 1)$$

$$\Rightarrow  \sqrt{3} \cos \theta - \cos \theta + \sqrt{3} \sin \theta - \sin \theta = \sqrt{3} \cos \theta + \cos \theta - \sqrt{3} \sin \theta - \sin \theta $$

$$\Rightarrow 2 \sqrt {3} \sin \theta = 2 \cos \theta $$

$$\Rightarrow \tan \theta = \dfrac {1}{\sqrt {3}} $$

$$\Rightarrow \theta = \tan^-1 \dfrac{1} {\sqrt {3}}$$

$$\Rightarrow \theta = 30^{\circ} $$

then $$ \sec \theta = \sec \30^\circ$$

                             = $$ \dfrac{2} {\sqrt {3}} $$ Ans 

Question 23 Show answer

From the top of a house A in a street, the angles of elevation and depression of the top and foot of another house B on the opposite side of the street are $$60^\circ$$ and $$45^\circ$$, respectively. If the height of house A is 36 m, then what is the height of house B? (Your answer should be nearest to an Integer.)

A

91 m

B

93 m

C

94 m

D

98 m

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Solution

From the above question, we draw the diagram is given below 

In $$ \triangle ADC $$

$$ \tan 45^\circ = \frac{DC} {AD} $$

$$ \Rightarrow 1 = \frac{36} {AD}$$

$$\Rightarrow AD = 36 m $$

In $$\triangle ADB $$

$$ \tan 60^\circ = \frac{BD}{AD}$$

$$ \Rightarrow BD= 36 \times \sqrt {3} $$ 

then Length of Building = BD + DC 

              $$ \Rightarrow 36  \sqrt{3} + 36 $$

             $$\Rightarrow 36 (1+\sqrt{3}) $$

             $$ \Rightarrow 36 \times (1+1.73) $$

             $$\Rightarrow 36 \times 2.73 $$

            $$\Rightarrow 98.28 $$

therefore Height of house B= 98 m Ans             

Question 24 Show answer

The value of $$\tan^2 48^\circ - \cosec^2 42^\circ + \cosec(67^\circ + \theta) - \sec(23^\circ - \theta)$$is:

A

-1

B

0

C

1

D

-2

View Explanation Hide Explanation

Solution

$$\tan^2 48^\circ - \cosec^2 42^\circ + \cosec(67^\circ + \theta) - \sec(23^\circ - \theta)$$

Now, we know that $$\cosec(90-\theta)=\sec\theta$$

So,$$\tan^2 48^\circ - \sec^2 (90^\circ-42^\circ) + \cosec(67^\circ + \theta) - \sec(23^\circ - \theta)$$

$$\Rightarrow \tan^2 (48^\circ) - \sec^2 (48^\circ) + \cosec(67^\circ + \theta) - \cosec(90^\circ-(23^\circ - \theta))$$

$$\Rightarrow \tan^2 (48^\circ) - \sec^2 (48^\circ) + \cosec(67^\circ + \theta) - \cosec(67^\circ - \theta)$$

$$\Rightarrow -1$$

Question 25 Show answer

From a point 12 m above the water level, the angle of elevation of the top of a hill is $$60^\circ$$ and the angle of depression of the base of the hill is $$30^\circ$$. What is the height (in m)of the hill?

A

36

B

$$48\sqrt{3}$$

C

48

D

$$36\sqrt{3}$$

Question 26 Show answer

If $$0 \leq \theta \leq 90^\circ$$, and $$\sec^{107} \theta + \cos^{107} \theta = 2$$, then $$(\sec \theta + \cos \theta)$$ is equal to:

A

$$2^{-107}$$

B

1

C

2

D

$$\frac{1}{2}$$

Question 27 Show answer

Two points A and B are on the ground and on opposite sides of a tower. A is closer to the foot of tower by 42 m than B. If the angles of elevation ofthe top of the tower, as observed from A and B are $$60^\circ$$ and $$45^\circ$$, respectively. then the height of the tower is closest to:

A

87.6 m

B

98.6 m

C

88.2 m

D

99.4 m

View Explanation Hide Explanation

Solution

Height of tower = OP

AB = 42 m 

$$tan45\degree = \frac{OP}{OA}$$

1 = OP/OA

OP = OA

$$tan60\degree = \frac{OP}{OA + AB}$$

$$\sqrt{3} = \frac{OP}{OP+ 42}$$

$$\sqrt{3} \times$$ (OP + 42) = OP

OP= $$\frac{42\sqrt{3}}{0.732} = \frac{72.744}{0.732}$$= 99.37 = 99.4 m

The height of tower is 99.4 m.

Question 28 Show answer

If $$\sin \theta - \cos \theta = 0, 0^\circ < \theta < 90^\circ$$, then the value of $$\sin^4 \theta + \cos^4 \theta$$ is:

A

$$\frac{1}{3}$$

B

$$1$$

C

$$\frac{1}{2}$$

D

$$\frac{1}{4}$$

View Explanation Hide Explanation

Solution

Given,

$$sin\theta-cos\theta=0$$

$$sin\theta=cos\theta$$

$$sin\theta=sin(90-\theta)$$

$$\theta=90-\theta$$

$$\theta+\theta=90$$

$$2\theta=90$$

$$\theta=45$$

$$sin^4\theta+cos^4\theta$$

$$sin^4(45)+cos^4(45)$$

$$(\dfrac{1}{\sqrt2})^4+(\dfrac{1}{\sqrt2}^4)$$

$$\dfrac{1}{4}+\dfrac{1}{4}$$

$$\dfrac{1}{2}$$

Question 29 Show answer

A pole stands vertically on a road, which goes in the north-south direction. P, Q are two points towards the north of the pole, such that PQ = b, and the angles ofelevation ofthe top ofthe pole at P, Q are a, B respectively. Then the height of the pole is:

A

$$\frac{b}{\tan \beta + \tan \alpha}$$

B

$$\frac{b}{\cot \beta - \cot \alpha}$$

C

$$\frac{b}{\tan \beta - \tan \alpha}$$

D

$$\frac{b \tan \alpha}{\tan \beta}$$

Question 30 Show answer

From the top of a lamp post of height x metres, two objects on the ground on the sameside of it (and in line with the foot of the lamp post) are observed at angles of depression of $$30^\circ$$ and $$60^\circ$$, respectively. The distance between the objects is $$32\sqrt{3}$$ m. The value of x is:

A

54

B

48

C

45

D

36

View Explanation Hide Explanation

Solution

from the question we draw the diagram 

Let BD =$$y $$

In the $$\triangle ABC $$ 

$$ \tan 30^\circ = \frac{x} {32\sqrt {3} +y} $$ 

$$ \frac {1} {\sqrt {3} } = \frac {x} {32 \sqrt {3} +y} $$ ...... Equestion (1)

then $$\triangle ABC $$

$$ \tan 60^\circ = \frac{x} {y} $$

$$ \sqrt {3} = \frac{x} {y} $$

$$ y = \frac {x} {\sqrt {3} } $$

Put the value $$ y $$ in Equation  (1)

$$ \frac {1} {\sqrt {3}} = \frac {x} { 32 \sqrt {3} + \frac {x} {\sqrt {3}}} $$

$$\Rightarrow 32 + \frac{x}{3} = x $$

$$ \Rightarrow 2 \frac {x}{3} = 32 $$

$$\Rightarrow x = 48  $$ Ans 

Question 31 Show answer

The angles of elevation of the top of a tower from two points on the ground at distances 32 m and 18 m from its base and in the same straight line with it are complementary. The height(in m) of the tower is .....

A

24

B

20

C

28

D

16

View Explanation Hide Explanation

Solution

Let $$ \angle ADB = \theta $$ 

then $$\angle ACD = (90^\circ - \theta) $$

In right triangle $$ \triangle ABD $$

$$ \tan \theta = \frac{AB}{BD} $$

$$\tan \theta = \frac{AB} {32} $$...........(1)

$$\triangle ABC $$ 

$$ \tan(90^\circ - \theta) = \frac{AB}{BC} $$

$$ \cot \theta = \frac{AB}{18} $$ ...........(2)

Multiplying Equation (1) and Equation (2)

$$ \tan \theta \times \frac{1}{\tan \theta} = \frac{AB}{32} \times \frac{AB}{18} $$

$$ 1 = \frac{(AB)^2} {32 \times 18} $$

$$ AB = \sqrt {18 \times 32 } $$

 $$ \Rightarrow AB = 2 \times 2 \times 2 \times 3 $$

$$ \Rightarrow AB = 24 $$ 

The height of tower = 24 m Ans 

Question 32 Show answer

If $$\sqrt{2} \sin(60^\circ - \alpha) = 1, 0^\circ < \alpha < 90^\circ$$, then $$\alpha$$ is equal to:

A

$$45^\circ$$

B

$$30^\circ$$

C

$$15^\circ$$

D

$$60^\circ$$

View Explanation Hide Explanation

Solution

As per the given question,

$$\sqrt(2) \sin(60^\circ - \alpha)= 1 $$ and $$0^\circ < \alpha < 90^\circ$$

$$\Rightarrow \sin(60^\circ - \alpha) = \dfrac{1}{\sqrt(2)}$$ 

$$\Rightarrow \sin(60^\circ - \alpha) = \sin(45^\circ) $$  (  put the value )

Taking the inverse of both side with $$\sin$$

$$\Rightarrow (60^\circ - \alpha) = 45^\circ $$ 

$$\Rightarrow \alpha = 60^\circ - 45^\circ $$

$$\alpha = 15^\circ $$ Ans

Question 33 Show answer

The value of $$\frac{1 - 2 \sin^2 \theta \cos^2 \theta}{\sin^4 \theta + \cos^4 \theta} - 1$$ is:

A

-1

B

1

C

$$-2 \sin^2 \theta \cos^2 \theta$$

D

0

View Explanation Hide Explanation

Solution

$$\frac{1 - 2 \sin^2 \theta \cos^2 \theta}{\sin^4 \theta + \cos^4 \theta} - 1$$

= $$\frac{1 - 2 \sin^2 \theta \cos^2 \theta - \sin^4 \theta - \cos^4 \theta}{\sin^4 \theta + \cos^4 \theta}$$

= $$\frac{1 - (sin^2 \theta + \cos^2 \theta)^2 }{\sin^4 \theta + \cos^4 \theta}$$

= $$\frac{1 - 1}{\sin^4 \theta + \cos^4 \theta}$$ = 0

Question 34 Show answer

The value of $$\frac{3(1 - 2 \sin^{2}x)}{\cos^{2}x - \sin^{2}x}$$ is:

A

4

B

3

C

1

D

2

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Solution

$$\frac{3(1-2sin^{2}x)}{cos^{2}x-sin^{2}x}$$

= $$\frac{3(1-2sin^{2}x)}{1-sin^{2}x-sin^{2}x}$$
($$\because sin^{2}x + cos^{2}x = 1$$)
= $$\frac{3(1-2sin^{2}x)}{1-2sin^{2}x}$$ = 3

Question 35 Show answer

Solve the following.
$$\frac{2 \sin 22^\circ}{\cos 68^\circ} - \frac{2 \cot 75^\circ}{5 \tan 15^\circ} - \frac{8 \tan 45^\circ \tan 20^\circ \tan 40^\circ \tan 50^\circ \tan 70^\circ}{5}$$

A

3

B

1

C

2

D

0

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Solution

$$\frac{2sin22^{0}}{cos68^{0}}-\frac{2cot75^{0}}{5tan15^{0}}-\frac{8tan45^{0}tan20^{0}tan40^{0}tan50^{0}tan70^{0}}{5}$$

$$\frac{2sin22^{0}}{cos(90 - 22)}-\frac{2cot(90 - 15)}{5tan15^{0}}-\frac{8tan45^{0}tan(90 - 70)tan(90 - 50)tan50^{0}tan70^{0}}{5}$$

$$\frac{2sin22^{0}}{sin22^{0}}-\frac{2tan15^{0}}{5tan15^{0}}-\frac{8tan45^{0}cot50^{0}cot70^{0}tan50^{0}tan70^{0}}{5}$$

$$2-\frac{2}{5}-\frac{8tan45^{0}}{5}$$

$$2-\frac{2}{5}-\frac{8}{5}$$ = 0

Question 36 Show answer

If $$\sec \theta + \tan \theta = p, 0^\circ < \theta < 90^\circ$$, then $$\frac{p^2 - 1}{p^2 + 1}$$ is equal to:

A

$$\cosec \theta$$

B

$$\sin \theta$$

C

$$\cos \theta$$

D

$$2 \cosec \theta$$

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Solution

$$\sec \theta + \tan \theta = p$$

$$(\sec \theta + \tan \theta)^2 = p^2$$ ---(1)

$$\frac{p^2 - 1}{p^2 + 1}$$

From eq(1),

= $$\frac{(\sec \theta + \tan \theta)^2 - 1}{(\sec \theta + \tan \theta)^2 + 1}$$

= $$\frac{\sec^2 \theta + \tan^2 \theta + 2\sec \theta \tan \theta - 1}{\sec^2 \theta + \tan^2 \theta + 2\sec \theta \tan \theta + 1}$$

= $$\frac{(\sec^2 \theta - 1) + \tan^2 \theta + 2\sec \theta \tan \theta}{\sec^2 \theta + 2\sec \theta \tan \theta + (\tan^2 \theta + 1)}$$

= $$\frac{2 \tan^2 \theta + 2\sec \theta \tan \theta }{2\sec^2 \theta + 2\sec \theta \tan \theta}$$

= $$\frac{\tan^2 \theta + \sec \theta \tan \theta }{\sec^2 \theta + \sec \theta \tan \theta}$$

= $$\frac{\tan \theta(\tan \theta + \sec \theta)}{\sec \theta (\sec \theta + \tan \theta)}$$

= $$\frac{\tan \theta}{\sec \theta}$$ = $$\sin\theta$$

Question 37 Show answer

If $$x = 4 \cos A + 5 \sin A$$  and $$y = 4 \sin A - 5 \cos A$$, then the value of $$x^2 + y^2$$ is:

A

25

B

41

C

0

D

16

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Solution

$$x = 4 \cos A + 5 \sin A$$

$$y = 4 \sin A - 5 \cos A$$

$$x^2 + y^2$$

= $$(4 \cos A + 5 \sin A)^2 + (4 \sin A - 5 \cos A)^2$$

($$(a + b)^2 = a^2 + b^2 + 2ab$$)

$$= (16 \cos^2 A + 25 \sin^2 A + 40\cos A\sin A) + (16 \sin^2 A + 25 \cos^2 A - 40\cos A\sin A)^2$$

($$ \sin^2 A + \cos^2 A = 1$$)

= 16 + 25 = 41

Question 38 Show answer

Solve the following.
$$\frac{\sin 40^\circ}{\cos 50^\circ} + \frac{\cosec 50^\circ}{\sec 40^\circ} - 4 \cos 50^\circ \cosec 40^\circ$$

A

2

B

-2

C

-1

D

1

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Solution

$$\frac{\sin 40^\circ}{\cos 50^\circ} + \frac{\cosec 50^\circ}{\sec 40^\circ} - 4 \cos 50^\circ \cosec 40^\circ$$

=$$\frac{\sin 40^\circ}{\cos(90^\circ - 40^\circ)} + \frac{\cosec(90^\circ - 40^\circ)}{\sec 40^\circ} - 4 \cos(90^\circ - 40^\circ) \cosec 40^\circ$$

=$$\frac{\sin 40^\circ}{\sin 40^\circ} + \frac{\sec 40^\circ}{\sec 40^\circ} - 4 \sin40^\circ \cosec 40^\circ$$

= 1 + 1 - 4 = -2

Question 39 Show answer

The value of $$(\cosec A + \cot A + 1)(\cosec A - \cot A + 1) - 2 \cosec A$$ is:

A

$$2 \cosec A$$

B

$$4 \cosec A$$

C

$$0$$

D

$$2$$

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Solution

$$(cosecA+cotA+1)(cosecA-cotA+1)-2cosecA$$ 

= $$(cosecA + 1)^2 - (cotA)^2 - 2cosecA$$

($$\because a^2 - b^2 = (a + b)(a - b)$$) 

= $$cosec^2A + 1 +  2cosecA - cot^2A - 2cosecA$$

($$cosec^2A  - cot^2A = 1)

= 1 + 1 = 2

Question 40 Show answer

If $$2 \sin \theta - 8 \cos^2 \theta + 5 = 0, 0^\circ < \theta < 90^\circ$$, then what is the value of $$(\tan 2\theta + \cosec 2\theta)?$$

A

$$3\sqrt{3}$$

B

$$\frac{4\sqrt{3}}{3}$$

C

$$2\sqrt{3}$$

D

$$\frac{5\sqrt{3}}{3}$$

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Solution

$$2 \sin \theta - 8 \cos^2 \theta + 5 = 0$$

$$\Rightarrow2 \sin \theta - 8(1 - \sin^2 \theta) + 5 = 0$$

$$\Rightarrow8 \sin^2 \theta + 2 \sin \theta - 3 = 0$$

$$\Rightarrow8 \sin^2 \theta + 6 \sin \theta- 4 \sin \theta- 3 = 0$$

$$\Rightarrow4\sin \theta(2\sin \theta - 1) + 3(2\sin \theta - 1) = 0$$

$$\Rightarrow(4\sin \theta + 3)(2\sin \theta - 1) = 0$$

$$\Rightarrow\sin \theta = 1/2$$

$$\Rightarrow\theta = 30\degree$$

$$(\tan 2\theta + \cosec 2\theta)$$

= $$(\tan 2 \times 30\degree+ \cosec 2 \times 30\degree)$$

= $$(\tan 60\degree+ \cosec 60\degree)$$

=$$\sqrt3 + \frac{2}{\sqrt3}$$

=$$ \frac{5}{\sqrt3} = \frac{5\sqrt3}{3}$$

Question 41 Show answer

If $$x \cos A - y \sin A = 1$$ and $$x \sin A + y \cos A = 4$$, then the value of $$17x^{2} + 17y^{2}$$ is:

A

0

B

7

C

49

D

289

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Solution

$$x \cos A - y \sin A=1$$

Take the square of both sides,

$$(x \cos A - y \sin A)^2=1$$

$$(x \cos A)^2 - 2xy \sin A \cos A + (y \sin A)^2 = 1$$ ---(1)

($$\because (a + b)^2 = a^2 + b^2 + 2ab$$)

$$x \sin A + y \cos A = 4$$

Take the square of both sides,

$$(x \sin A + y \cos A)^2=16$$

$$(x \sin A)^2 + 2xy \sin A \cos A + (y \cos A)^2 =16$$ ---(2)

Addition of eq(1) and eq(2),

$$(x \cos A)^2 + (x \sin A)^2 + (y \sin A)^2 + (y \cos A)^2 = 17$$

$$x^2 + y^2 = 17$$

$$17x^{2}+17y^{2} = 17 \times 17$$

$$17x^{2}+17y^{2} = 289$$

Question 42 Show answer

if $$0 < A, B < 45^\circ, \cos(A + B) = \frac{24}{25}$$ and $$\sin(A - B) = \frac{15}{17}$$, then $$\tan 2A$$ is

A

1

B

0

C

$$\frac{416}{87}$$

D

$$\frac{213}{4}$$

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Solution

$$\tan 2A$$

= tan((A + B) + (A - B))

=$$\frac{tan(A + B) + tan(A + B)}{1 - tan(A + B)tan(A + B)}$$ ---(1)

$$(\because tan(a + b) = \frac{tana + tanb}{1 - tana.tanb})$$

tan(A + B) = $$\frac{sin(A + B)}{cos(A - B)}$$

tan(A + B) = $$\frac{\sqrt{1 - cos^2(A + B)}}{cos(A + B)}$$ 

tan(A + B) = $$\frac{\sqrt{1 - (24/25)^2}}{(24/25)}$$

tan(A + B) = $$\frac{\sqrt{49/25}}{(24/25)}$$ = 7/24

tan(A - B) = $$\frac{sin(A - B)}{cos(A - B)}$$

tan(A - B)= $$\frac{sin(A - B)}{\sqrt{1 - sin^2(A - B)}}$$

tan(A - B) = $$\frac{15/17}{\sqrt{1 - (15/17)^2}}$$

tan(A - B) = $$\frac{15/17}{\sqrt{64/17)^2}}$$ = 15/8

From eq(1),

=$$\frac{\frac{7}{24} +\frac{15}{8}}{1 - \frac{7}{24}.\frac{15}{8}}$$

=$$\frac{416}{87}$$

Question 43 Show answer

The value of $$\frac{\tan 30^\circ \cosec 60^\circ + \tan 60^\circ \sec 30^\circ}{\sin^2 30^\circ + 4 \cot^2 45^\circ - \sec^2 60^\circ}$$ is:

A

$$\frac{2}{3}$$

B

$$\frac{32}{99}$$

C

$$\frac{8}{3}$$

D

$$\frac{32}{3}$$

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Solution

$$\frac{\tan 30^\circ \cosec 60^\circ + \tan 60^\circ \sec 30^\circ}{\sin^2 30^\circ + 4 \cot^2 45^\circ - \sec^2 60^\circ}$$

On put the values,

= $$\frac{\frac{1}{\sqrt3} \times \frac{2}{\sqrt3} + \sqrt3 \times \frac{2}{\sqrt3}}{(\frac{1}{2})^2 + 4 \times 1 - 2^2}$$

= $$\frac{\frac{2}{3} + 2}{\frac{1}{4} + 4  - 4}$$

= $$\frac{\frac{8}{3}}{\frac{1}{4}}$$

= $$\frac{32}{3}$$

Question 44 Show answer

If $$\tan \theta - \cot \theta = \cosec \theta, 0^\circ < \theta < 90^\circ$$, then what is the value of $$\frac{2 \tan \theta - \cos \theta}{\sqrt{3} \cot \theta + \sec \theta}$$?

A

$$\frac{3\sqrt{3} - 1}{6}$$

B

$$\frac{4\sqrt{3} - 1}{6}$$

C

$$\frac{2\sqrt{3} - 1}{3}$$

D

$$\frac{2(2\sqrt{3} - 1)}{3}$$

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Solution

$$\tan \theta - \cot \theta = \cosec \theta$$

$$\tan \theta - \frac{1}{tan \theta} = \cosec \theta$$

$$\tan^2 \theta - 1 = \cosec \theta\tan \theta$$

$$\sec^2 \theta - 1 - 1 = \sec \theta$$

$$\sec^2 \theta - \sec \theta - 2 = 0$$

$$\sec^2 \theta - 2\sec \theta + \sec \theta - 2 = 0$$

$$\sec \theta(\sec \theta - 2) +1(\sec \theta - 2) = 0$$

$$ (\sec \theta + 1)(\sec \theta - 2) = 0$$

For $$0^\circ < \theta < 90^\circ$$,

$$\sec \theta$$ = 2

$$\cos \theta$$ = 1/2

$$\theta = 60\degree$$

Now,

$$\frac{2 \tan \theta - \cos \theta}{\sqrt{3} \cot \theta + \sec \theta}$$

Put the value of $$\theta$$,

= $$\frac{2 \tan 60\degree - \cos60\degree}{\sqrt{3} \cot 60\degree + \sec 60\degree}$$

= $$\frac{2\sqrt3- \frac{1}{2}}{\sqrt{3} \times \frac{1}{\sqrt3}+ 2}$$

= $$\frac{4\sqrt3- 1}{6}$$

Question 45 Show answer

The value of $$4\left[\frac{(1 - \sec A)^{2} + (1 + \sec A)^{2}}{1 + \sec^{2}A}\right]$$ is :

A

4

B

1

C

2

D

8

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Solution

$$4\left[\frac{(1 - \sec A)^{2} + (1 + \sec A)^{2}}{1 + \sec^{2}A}\right]$$

= $$4\left[\frac{1 + \sec^{2}A  - 2\sec A + 1 + \sec^{2}A + 2\sec A}{1 + \sec^{2}A}\right]$$

= $$4\left[\frac{2(1 + \sec^{2}A) }{1 + \sec^{2}A}\right]$$ = 8

Question 46 Show answer

The value of $$\frac{\tan 30^\circ + \tan 60^\circ}{\cos 30^\circ}$$ is:

A

$$1 + \sqrt{3}$$

B

$$\frac{8}{\sqrt{3}}$$

C

$$\frac{8}{3}$$

D

$$\sqrt{3} + 3$$

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Solution

$$\frac{\tan 30^\circ + \tan 60^\circ}{\cos 30^\circ}$$

put the value in the above expression,

= $$\frac{\frac{1}{\sqrt{3}} + \sqrt{3}}{\frac{\sqrt{3}}{2}}$$ 

= $$\frac{1+3}{\sqrt{3}} \times$$ $$2/\sqrt3$$

= $$\frac{8}{3}$$

Question 47 Show answer

Whatis the value of $$\frac{\cosec(78^\circ + \theta) - \sec(12^\circ - \theta) - \tan(67^\circ + \theta) + \cot(23^\circ - \theta)}{\tan 13^\circ \tan37^\circ \tan45^\circ \tan53^\circ \tan77^\circ}$$

A

1

B

0

C

2

D

-1

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Solution

$$\frac{\cosec(78^\circ + \theta) - \sec(12^\circ - \theta) - \tan(67^\circ + \theta) + \cot(23^\circ - \theta)}{\tan 13^\circ \tan37^\circ \tan45^\circ \tan53^\circ \tan77^\circ}$$

$$\frac{\cosec(90^\circ - (12^\circ - \theta)) - \sec(12^\circ - \theta) - \tan(90^\circ -  (23^\circ - \theta)) + \cot(23^\circ - \theta)}{\tan 13^\circ \tan37^\circ \tan45^\circ \tan53^\circ \tan77^\circ}$$

$$\frac{\sec(12^\circ - \theta) - \sec(12^\circ - \theta) - \cot(23^\circ - \theta) + \cot(23^\circ - \theta)}{\tan 13^\circ \tan37^\circ \tan45^\circ \tan53^\circ \tan77^\circ}$$

$$\frac{0}{\tan 13^\circ \tan37^\circ \tan45^\circ \tan53^\circ \tan77^\circ}$$ = 0

Question 48 Show answer

If $$2 \sin \theta + 15 \cos^2 \theta = 7, 0^\circ < \theta < 90^\circ$$, then $$\tan \theta + \cos \theta + \sec \theta =$$

A

$$3\frac{3}{5}$$

B

3

C

$$3\frac{4}{5}$$

D

4

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Solution

$$2 \sin \theta + 15 \cos^2 \theta = 7$$

$$2 \sin \theta + 15 (1 - sin^2 \theta) = 7$$

$$2 \sin \theta + 15 - 15 sin^2 \theta = 7$$

$$2 \sin \theta + 8 - 15 sin^2 \theta = 0$$

$$15 sin^2 \theta - 2 \sin \theta - 8 = 0 $$

$$sin^2 \theta - \frac{2\sin \theta}{15} - \frac{8}{15} = 0 $$

$$sin^2 \theta + \frac{2\sin \theta}{3} - \frac{4\sin \theta}{5} - \frac{8}{15} = 0 $$

$$sin \theta(sin \theta + \frac{2}{3}) - \frac{4}{5}(sin \theta + \frac{2}{3}) = 0 $$

$$(sin \theta - \frac{4}{5})(sin \theta + \frac{2}{3}) = 0$$

For  $$0^\circ < \theta < 90^\circ$$,

$$sin \theta = \frac{4}{5}$$

Perpendicular = 4

Hypotenuses = 5

By triplet 3-4-5,

Base = 3 

Now,

$$\tan \theta + \cos \theta + \sec \theta$$

= $$\frac{4}{3} + \frac{3}{5} +\frac{5}{3}$$

= $$\frac{20 + 9 + 25}{15}$$

= $$\frac{54}{15}$$ = $$\frac{18}{5}$$

= $$3\frac{3}{5}$$

Question 49 Show answer

If $$7 \sin^2 \theta - \cos^2 \theta + 2 \sin \theta = 2, 0^\circ < \theta < 90^\circ$$, then the value of $$\frac{\sec 2\theta + \cot 2\theta}{\cosec 2\theta + \tan 2\theta}$$ is:

A

$$\frac{2\sqrt{3} + 1}{3}$$

B

1

C

$$\frac{1}{5}(1 + 2\sqrt{3})$$

D

$$\frac{2}{5}(1 + \sqrt{3})$$

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Solution

$$7 \sin^2 \theta - \cos^2 \theta + 2 \sin \theta = 2$$

$$\Rightarrow$$ $$7 \sin^{2} \theta - \cos^{2} \theta + 2 \sin \theta - 2 = 0$$

$$\Rightarrow$$ $$7 \sin^{2} \theta - (1 - \sin^{2} \theta) + 2 \sin \theta - 2 = 0$$

$$\Rightarrow$$ $$8 \sin^{2} \theta + 2 \sin \theta - 3 = 0$$

$$\Rightarrow$$ $$8 \sin^{2} \theta + 6 \sin \theta - 4\sin \theta- 3 = 0$$

$$\Rightarrow$$ $$2\sin \theta(4 \sin\theta + 3) -1(4 \sin\theta + 3) = 0$$

$$\Rightarrow$$ $$(2\sin \theta - 1)(4 \sin\theta + 3) = 0$$

For $$ 0^\circ < \theta < 90^\circ$$,

$$\sin \theta = 1/2$$

$$\theta = 30\degree$$

$$\frac{\sec 2 \times 30+ \cot 2\times 30}{\cosec 2\times 30 + \tan 2\times 30}$$

$$\Rightarrow$$ $$\frac{\sec 60 + \cot 60}{\cosec 60 + \tan 60}$$ 
$$\Rightarrow$$ $$\frac{2+ \frac{1}{\sqrt3}}{\frac{2}{\sqrt3} + \sqrt3}$$

$$\Rightarrow$$ $$\frac{2\sqrt3 + 1}{2+ 3}$$

$$\Rightarrow$$ $$\frac{2\sqrt3 + 1}{5}$$

Question 50 Show answer

In the figure, what is the value of $$\cot \theta ?$$

A

$$\frac{17}{18}$$

B

$$\frac{15}{17}$$

C

$$\frac{15}{8}$$

D

$$\frac{8}{15}$$

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Solution

From the triangle,

PQ$$^2$$ + QR$$^2$$ = PR$$^2$$

$$\Rightarrow$$  8$$^2$$ + QR$$^2$$ = 17$$^2$$

$$\Rightarrow$$  64 + QR$$^2$$ = 289

$$\Rightarrow$$  QR$$^2$$ = 225

$$\Rightarrow$$  QR = 15 cm

$$\therefore\ $$ $$\cot\theta\ =\frac{\text{Adjacent side}}{\text{Opposite side}}=\frac{QR}{PQ}=\frac{15}{8}$$

Hence, the correct answer is Option C

Question 51 Show answer

The value of $$\cos 0^\circ \cos 30^\circ \cos 45^\circ \cos 60^\circ \cos 90^\circ$$ is:

A

5

B

3

C

0

D

$$\frac{\sqrt{6}}{8}$$

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Solution

$$\cos 0^\circ \cos 30^\circ \cos 45^\circ \cos 60^\circ \cos 90^\circ$$ = 0

($$\because \cos 0^\circ = 0$$)

Question 52 Show answer

What is the value of $$\sin 30^\circ + \cos 30^\circ - \tan 45^\circ $$?

A

$$\frac{\sqrt{3} - 1}{2}$$

B

$$\frac{\sqrt{2} + 1}{\sqrt{2}}$$

C

$$\frac{\sqrt{3} + 1}{2}$$

D

$$\frac{1 - \sqrt{3}}{2}$$

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Solution

$$\sin 30^\circ + \cos 30^\circ - \tan 45^\circ $$

= $$1/2 + \frac{\sqrt3}{2} - 1$$

= $$\frac{1 + \sqrt3 - 2}{2}$$

= $$\frac{\sqrt3 - 1}{2}$$

Question 53 Show answer

If $$\cot \theta + \tan\theta = 2 \sec \theta, 0^\circ < \theta < 90^\circ $$, then the value of $$\frac{\tan 2 \theta - \sec \theta}{\cot 2 \theta + \cosec \theta}$$

A

$$\frac{2\sqrt{3}-1}{5}$$

B

$$\frac{3-\sqrt{2}}{5}$$

C

$$\frac{2\sqrt{3}-1}{11}$$

D

$$\frac{3-\sqrt{2}}{11}$$

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Solution

$$\cot \theta + \tan\theta = 2 \sec \theta$$

On putting the value of $$\theta$$ = 30$$\degree$$

$$\cot30\degree + \tan30\degree = 2 \sec30\degree$$

$$ \sqrt3 + \frac{1}{\sqrt3} = 2 \times \frac{2}{\sqrt3}$$

$$\frac{4}{\sqrt3} = \frac{4}{\sqrt3}$$

$$\frac{\tan 2 \theta - \sec \theta}{\cot 2 \theta + \cosec \theta}$$

= $$\frac{\tan 2 \times 30\degree - \sec30\degree }{\cot 2 \times 30\degree + \cosec30\degree }$$

= $$\frac{\sqrt{\ 3}-\frac{2}{\sqrt{\ 3}}}{\frac{1}{\sqrt{\ 3}}+2}$$

= $$\frac{\frac{1}{\sqrt{\ 3}}}{\frac{1 + 2\sqrt3}{\sqrt{\ 3}}}$$

= $$\frac{1}{1 + 2\sqrt3} \times \frac{1 - 2\sqrt3}{1 - 2\sqrt3}$$

= $$\frac{1}{1 + 2\sqrt3} \times \frac{1 - 2\sqrt3}{1^2 - (2\sqrt3)^2}$$

= $$\frac{2\sqrt3 - 1}{11}$$

Question 54 Show answer

If $$5 \cos \theta - 12 \sin \theta = 0$$, then what is the value of $$\frac{1 + \sin \theta + \cos \theta}{1 - \sin \theta + \cos \theta}$$

A

$$\dfrac{5}{4}$$

B

$$\dfrac{3}{2}$$

C

$$\dfrac{3}{4}$$

D

$$\dfrac{5}{2}$$

View Explanation Hide Explanation Video Solution

Solution

$$5 \cos \theta - 12 \sin \theta = 0$$

$$tan \theta = \frac{5}{12}$$

We know that $$tan \theta = \dfrac{perpendicular}{base}$$ so,

By the triplet 5-12-13,

Hypotenuse = 13

$$sin \theta = \dfrac{5}{13}$$

$$cos \theta = \dfrac{12}{13}$$

$$\dfrac{1 + \sin \theta + \cos \theta}{1 - \sin \theta + \cos \theta}$$

= $$\dfrac{1 + \dfrac{5}{13} + \dfrac{12}{13}}{1 -  \dfrac{5}{13} + \dfrac{12}{13}}$$

= $$\dfrac{30}{20} = \dfrac{3}{2}$$

Question 55 Show answer

If $$5 \sin^2 \theta + 14 \cos \theta = 13, 0^\circ < \theta < 90^\circ$$, then what is the value of $$\frac{\sec \theta + \cot \theta}{\cosec \theta + \tan \theta}$$?

A

$$\frac{40}{29}$$

B

$$\frac{32}{27}$$

C

$$\frac{21}{28}$$

D

$$\frac{9}{8}$$

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Solution

$$5 \sin^2 \theta + 14 \cos \theta = 13$$

$$5(1 - cos^2 \theta)+ 14 \cos \theta = 13$$

$$-5cos^2 \theta + 14 \cos \theta = 8$$

$$5cos^2 \theta - 10 \cos \theta - 4 \cos \theta + 8 = 0$$

$$5cos \theta(\cos \theta - 2)  - 4 (\cos \theta - 2) = 0$$

$$(5cos \theta - 4)(\cos \theta - 2) = 0$$

$$(\cos \theta - 2) $$(it is not possible)

For $$0^\circ < \theta < 90^\circ$$,

$$(5cos \theta - 4) = 0$$

$$cos \theta = 4/5$$

by triplet 3-4-5,

$$\sin \theta = 3/5$$

Now,

$$\dfrac{\sec \theta + \cot \theta}{\cosec \theta + \tan \theta}$$

= $$\dfrac{\sec \theta(1 + \frac{1}{\sin \theta})}{\sec \theta(\cosec \theta\cos \theta + \sin \theta)}$$

= $$\dfrac{1 + \frac{1}{\sin \theta}}{\cosec \theta\cos \theta + \sin \theta}$$

= $$\dfrac{1 + \frac{5}{3}}{\frac{5}{3} \times \frac{4}{5} + \frac{3}{5}}$$

= $$\dfrac{\frac{8}{3}}{ \frac{4}{3} + \frac{3}{5}}$$
= $$\dfrac{\frac{8}{3}}{\frac{29}{15}}$$ = $$\dfrac{40}{29}$$

Question 56 Show answer

If $$\frac{\sec \theta - \tan \theta}{\sec \theta + \tan \theta} = \frac{3}{5}$$, then the value of $$\frac{\cosec \theta + \cot \theta}{\cosec \theta - \cot \theta}$$ is:

A

$$24 + \sqrt{15}$$

B

$$31 + 8\sqrt{15}$$

C

$$27 + \sqrt{15}$$

D

$$33 + 4\sqrt{15}$$

View Explanation Hide Explanation

Solution

$$\frac{\sec \theta - \tan \theta}{\sec \theta + \tan \theta} = \frac{3}{5}$$

Divide numerator and den. by tan$$\theta$$,

 $$\frac{\cosec \theta - 1}{\cosec \theta + 1} = \frac{3}{5}$$

Using componendo dividendo rule,

(if $$\frac{p + q}{p - q} = \frac{a}{b} then \frac{p}{q} = \frac{a + b}{a - b}$$)

cosec$$\theta = \frac{5 + 3}{5 - 3} = 4

$$cot^2θ=cosec^2θ−1$$

$$\therefore \cot^2 \theta = 4^2 - 1$$

$$\therefore \cot \theta = \sqrt {15}$$

$$\frac{cosec \theta + cot \theta}{cosec \theta - cot \theta} = \frac{4 + \sqrt{15}}{4 - \sqrt{15}}​​$$

Rationalizing, we get

= $$\frac{4 + \sqrt {15}}{4 - \sqrt {15}} \times \frac{4 + \sqrt {15}}{4 + \sqrt {15}}$$

= 31 + 8$$\sqrt{15}$$ 

Question 57 Show answer

The expression $$3 \sec^2 \theta \tan^2 \theta + \tan^6 \theta - \sec^6 \theta$$ is equal to:

A

-2

B

1

C

2

D

-1

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Solution

$$3 \sec^2 \theta \tan^2 \theta + \tan^6 \theta - \sec^6 \theta$$

= $$-3 \sec^2 \theta \tan^2 \theta(\tan^2 \theta - \sec^2 \theta) + \tan^6 \theta - \sec^6 \theta$$

($$\because \tan^2 \theta - \sec^2 \theta = -1$$)

= ($$\tan^2 \theta - \sec^2 \theta$$)^3

($$\because(a - b)^3 = a^3 - b^3 - 3ab(a - b)$$)

= $$(-1)^3$$ = -1

Question 58 Show answer

The value of $$\frac{\sin 30^\circ \sin 60^\circ}{\cos 60^\circ \cos 30^\circ } - \tan 45^\circ$$ is:

A

5

B

0

C

$$\frac{2-\sqrt{2}}{2}$$

D

2

View Explanation Hide Explanation

Solution

$$\frac{\sin 30^\circ \sin 60^\circ}{\cos 60^\circ \cos 30^\circ } - \tan 45^\circ$$

Put the values in the expression,

$$\frac{\frac{1}{2} \times \frac{\sqrt3}{2}}{\frac{1}{2} \times  \frac{\sqrt3}{2} } - 1$$

= 1 - 1 = 0

Question 59 Show answer

In the given figure, $$\cos\theta$$ is equal to:

A

$$\frac{5}{13}$$

B

$$\frac{12}{13}$$

C

$$\frac{5}{12}$$

D

$$\frac{12}{5}$$

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Solution

In $$\triangle$$PQR,

QR$$^2$$ + PR$$^2$$ = PQ$$^2$$

$$\Rightarrow$$  12$$^2$$ + PR$$^2$$ = 13$$^2$$

$$\Rightarrow$$  144 + PR$$^2$$ = 169

$$\Rightarrow$$  PR$$^2$$ = 25

$$\Rightarrow$$  PR = 5 units

$$\therefore\ $$ $$\cos\theta=\frac{\text{adjacent side}}{\text{hypotenuse}}=\frac{\text{PR}}{\text{PQ}}=\frac{5}{13}$$

Hence, the correct answer is Option A

Question 60 Show answer

If $$11 \sin^{2} \theta - \cos^{2} \theta + 4 \sin \theta - 4 = 0, 0^\circ < \theta < 90^\circ$$, then what is the value of $$\frac{\cos 2\theta + \cot 2 \theta}{\sec 2 \theta - \tan 2 \theta}$$

A

$$\frac{12+7\sqrt{3}}{6}$$

B

$$\frac{12+5\sqrt{3}}{3}$$

C

$$\frac{10+5\sqrt{3}}{3}$$

D

$$\frac{10+7\sqrt{3}}{6}$$

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Solution

$$11 \sin^{2} \theta - \cos^{2} \theta + 4 \sin \theta - 4 = 0$$

$$11 \sin^{2} \theta - (1 - \sin^{2} \theta) + 4 \sin \theta - 4 = 0$$

$$12 \sin^{2} \theta + 4 \sin \theta - 5 = 0$$

$$12 \sin^{2} \theta + 10 \sin \theta - 6\sin \theta- 5 = 0$$

$$2\sin \theta(6 \sin\theta + 5) -1(6 \sin\theta + 5) = 0$$

$$(2\sin \theta - 1)(6 \sin\theta + 5) = 0$$

For $$ 0^\circ < \theta < 90^\circ$$,

$$\sin \theta = 1/2$$

$$\theta = 30\degree$$

$$\frac{\cos 2\theta + \cot 2 \theta}{\sec 2 \theta - \tan 2 \theta}$$

On putting the value of $$\theta$$,

$$\frac{\cos 2\times 30 + \cot 2 \times 30}{\sec 2 \times 30 - \tan 2 \times 30}$$

$$\frac{\cos 60 + \cot 60}{\sec60 - \tan 60}$$

$$\frac{\frac{1}{2} + \frac{1}{\sqrt3}}{2 - \sqrt3}$$

$$\frac{2 + \sqrt3}{2\sqrt3(2 - \sqrt3)}$$

$$\frac{2 + \sqrt3}{4\sqrt3- 6}$$

$$\frac{2 + \sqrt3}{4\sqrt3- 6} \times \frac{4\sqrt3 + 6}{4\sqrt3 + 6}$$

$$\frac{(2 + \sqrt3)(4\sqrt3 + 6)}{(4\sqrt3)^2- 6^2} $$

$$\frac{8\sqrt3 + 12 + 12 + 6\sqrt3}{12} $$

$$\frac{12 + 7\sqrt3}{6} $$

Question 61 Show answer

If $$(2 \sin A + \cosec A) = 2 \sqrt{2}$$, $$0^\circ < A < 90^\circ$$ then the value of $$2(\sin^{4}A + \cos^{4}A)$$ is:

A

2

B

1

C

4

D

0

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Solution

$$(2 \sin A + \cosec A) = 2 \sqrt{2}$$

To find the value A, we satisfy the above equation so put the value of A = 45$$\degree$$

$$(2 \sin 45 \degree + \cosec 45 \degree) = 2 \sqrt{2}$$

$$(2 \times \frac{1}{\sqrt{2}} + \sqrt{2}) = 2 \sqrt{2}$$

$$ 2\sqrt{2} = 2\sqrt{2}$$

$$2(\sin^{4}A + \cos^{4}A)$$

= $$2(\sin^{4}45 \degree + \cos^{4}45\degree)$$

= $$2((\frac{1}{\sqrt{2}})^{4} + (\frac{1}{\sqrt{2}})^{4})$$

= $$2(\frac{1}{4} + \frac{1}{4}) = 2(\frac{1}{2}) = 1$$

Question 62 Show answer

If $$3 \sec^2 \theta + \tan \theta = 7, 0^\circ < \theta <90^\circ$$, then the value of $$\frac{\cosec 2 \theta + \cos \theta}{\sin 2 \theta + \cot \theta}$$ is:

A

$$\frac{2 + 3\sqrt{2}}{4}$$

B

$$\frac{3 + \sqrt{2}}{2}$$

C

$$\frac{2 + \sqrt{2}}{4}$$

D

$$\frac{2 + \sqrt{3}}{2}$$

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Solution

$$3 \sec^2 \theta + \tan \theta = 7$$

$$3 (1 + tan^2\theta) + \tan \theta = 7$$

$$4tan^2\theta = 4$$

$$tan^2\theta = 1$$

$$\theta = 45\degree$$

$$\frac{\cosec 2 \theta + \cos \theta}{\sin 2 \theta + \cot \theta}$$

= $$\frac{\cosec 2 \times 45\degree + \cos 45\degree}{\sin 2 \times 45\degree + \cot 45\degree}$$

= $$\frac{\cosec 90\degree + \cos 45\degree}{\sin 90\degree + \cot 45\degree}$$

Put the value of $$\theta$$,

= $$\frac{1 + \frac{1}{\sqrt{2}}}{1 + 1}$$

= $$\frac{\sqrt{2} + 1}{2\sqrt{2}}$$

= $$\frac{\sqrt{2} + 1}{2\sqrt{2}}$$ $$\times \frac{\sqrt{2}}{\sqrt{2}}$$

= $$\frac{\sqrt{2} + 2}{4}$$

Question 63 Show answer

The value of $$\cos 10^\circ \cos 30^\circ \cos 50^\circ \cos 70^\circ \cos 90^\circ$$ is:

A

3

B

0

C

5

D

1

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Solution

$$cos10^{0}cos30^{0}cos50^{0}cos70^{0}cos90^{0}$$

= $$cos10^{0}cos30^{0}cos50^{0}cos70^{0}\times 0$$

($$\because cos90^{0} = 0$$)

= 0

Question 64 Show answer

If $$12 \cos^2 \theta - 2 \sin^2 \theta + 3\cos \theta = 3, 0^\circ < \theta < 90^\circ$$, then what is the value of $$\frac{\cosec \theta + \sec \theta}{\tan \theta + \cot \theta}$$?

A

$$\frac{4 + \sqrt{3}}{4}$$

B

$$\frac{1 + 2\sqrt{2}}{2}$$

C

$$\frac{1 + \sqrt{3}}{2}$$

D

$$\frac{2 + \sqrt{3}}{4}$$

View Explanation Hide Explanation

Solution

$$12 \cos^2 \theta - 2 \sin^2 \theta + 3\cos \theta = 3$$

$$12 \cos^2 \theta - 2(1 -  \cos^2 \theta) + 3\cos \theta = 3$$

$$14 \cos^2 \theta + 3\cos \theta = 5$$

Put the value of $$\theta = 60\degree$$,

$$14 \cos^2 60\degree + 3\cos 60\degree = 5$$

$$14 \times \frac{1}{2} + 3 \times \frac{1}{2} = 5$$

5 = 5

L.H.S. = R.H.S.

$$\frac{\cosec \theta + \sec \theta}{\tan \theta + \cot \theta}$$

= $$\frac{\cosec 60\degree + \sec 60\degree}{\tan 60\degree + \cot 60\degree}$$

= $$\frac{\frac{2}{\sqrt3} + 2}{\sqrt3 + \frac{1}{\sqrt3}}$$

= $$\frac{\frac{2 + 2\sqrt3}{\sqrt3}}{\frac{3 + 1}{\sqrt3}}$$

= $$\frac{1 + \sqrt3}{2}$$

Question 65 Show answer

The value of $$(\cosec 30^\circ - \tan 45^\circ) \cot 60^\circ \tan 30^\circ$$ is:

A

$$\frac{1}{3}$$

B

$$1$$

C

$$3$$

D

$$\frac{1}{\sqrt{3}}$$

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Solution

$$(\cosec 30^\circ - \tan 45^\circ) \cot 60^\circ \tan 30^\circ$$

On put the value,

= $$(2 - 1)\frac{1}{\sqrt3} \times \frac{1}{\sqrt3}$$

= 1/3

Question 66 Show answer

If $$A + B = 45^\circ$$, then the value of $$2(1 + \tan A)(1 + \tan B)$$ is:

A

4

B

0

C

2

D

1

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Solution

$$A + B = 45^\circ$$

On taking tan both sides,

$$tan(A + B) = tan(45^\circ)$$

$$\frac{\tan A + \tan B}{1 - \tan A \tan B} = 1$$

$$\tan A + \tan B = 1 - \tan A \tan B$$

$$\tan A + \tan B + \tan A \tan B = 1$$ ---(1)

Now,

$$2(1 + \tan A)(1 + \tan B)$$

= $$2(1 + \tan A + \tan B + \tan A \tan B)$$

from the eq(1),

= $$2(1 + 1)$$

= 4

Question 67 Show answer

If $$(\cos^{2} \theta - 1)(1 + \tan^{2} \theta) + 2 \tan^2 \theta = 1, 0^\circ < \theta < 90^\circ$$ then $$\theta$$ is:

A

$$ 30^\circ $$

B

$$ 90^\circ $$

C

$$ 60^\circ $$

D

$$ 45^\circ $$

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Solution

$$(\cos^{2} \theta - 1)(1 + \tan^{2} \theta) + 2 \tan^2 \theta = 1$$
 $$(-\sin^{2} \theta )(sec^2 \theta) + 2 \tan^2 \theta = 1$$

$$\frac{-\sin^{2} \theta}{cos^2 \theta} + 2 \tan^2 \theta = 1$$

$$-\tan^2 \theta + 2 \tan^2 \theta = 1$$

$$\tan^2 \theta = 1$$

For 0^\circ < \theta < 90^\circ$$,

$$\theta = 45\degree$$

Question 68 Show answer

Solve the following.
$$\left(\frac{\sin 27^\circ}{\cos 63^\circ}\right) - \left(\frac{\cos 27^\circ}{\sin 63^\circ}\right)^{2}$$

A

2

B

0

C

-1

D

1

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Solution

$$\left(\frac{\sin 27^\circ}{\cos 63^\circ}\right) - \left(\frac{\cos 27^\circ}{\sin 63^\circ}\right)^{2}$$

$$\left(\frac{\sin(90^\circ - 63^\circ)}{\cos 63^\circ}\right) - \left(\frac{\cos(90^\circ - 63^\circ)}{\sin 63^\circ}\right)^{2}$$

= $$\left(\frac{\\cos 63^\circ}{\cos 63^\circ}\right) - \left(\frac{\sin 63^\circ}{\sin 63^\circ}\right)^{2}$$

= 1 - 1 = 0

Question 69 Show answer

The value of $$\frac{\sec^6 \theta - \tan^6 \theta - 3\sec^2 \theta \tan^2 \theta + 1}{\cos^4 \theta - \sin^4 \theta + 2 \sin^2 \theta + 2}$$ is:

A

$$\frac{3}{4}$$

B

$$\frac{2}{3}$$

C

$$\frac{1}{2}$$

D

$$1$$

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Solution

$$\frac{\sec^6 \theta - \tan^6 \theta - 3\sec^2 \theta \tan^2 \theta + 1}{\cos^4 \theta - \sin^4 \theta + 2 \sin^2 \theta + 2}$$

= $$\frac{\sec^6 \theta - \tan^6 \theta - 3\sec^2 \theta \tan^2 \theta(\sec^2 \theta - \tan^2 \theta) + 1}{(\cos^2 \theta - \sin^2 \theta)(\cos^2 \theta + \sin^2 \theta) + 2 \sin^2 \theta + 2}$$

($$(a - b)^3 = a^3 - b^3 - 3ab(a - b)$$)

($$a^2 - b^2 = (a+b)(a-b)$$)

= $$\frac{(\sec^6 \theta - \tan^6 \theta)^3 + 1}{\cos^2 \theta - \sin^2 \theta + 2 \sin^2 \theta + 2}$$

= $$\frac{1 + 1}{\cos^2 \theta + \sin^2 \theta + 2}$$

= $$\frac{2}{1 + 2}$$

= $$\frac{2}{3}$$

Question 70 Show answer

If $$5 \sin \theta = 4$$, then the value of $$\frac{\sec \theta + 4 \cot \theta}{4 \tan \theta - 5 \cos \theta}$$ is:

A

$$\frac{3}{2}$$

B

$$1$$

C

$$\frac{5}{4}$$

D

$$2$$

View Explanation Hide Explanation

Solution

$$5 \sin \theta = 4$$

$$\sin \theta = 4/5$$

$$\frac{perpendicular}{hypotenuses} = \frac{4}{5}$$

By triplet 3-4-5,

Base = 3

$$cos\theta = base/hypotenuses = 3/5$$

$$tan\theta = perpendicular/base = 4/3$$

$$\frac{\sec \theta + 4 \cot \theta}{4 \tan \theta - 5 \cos \theta}$$

= $$\frac{\frac{1}{\cos \theta} +  \frac{4}{\tan \theta}}{4 \tan \theta - 5 \cos \theta}$$

= $$\frac{\frac{1}{3/5} + \frac{4}{4/3}}{4 \times 4/3 - 5 \times 3/5}$$

= $$\frac{\frac{5}{3} + 3}{4 \times 4/3 - 5 \times 3/5}$$

= $$\frac{\frac{14}{3}}{\frac{16}{3} - 3}$$

= $$\frac{14}{7}$$ = 2

Question 71 Show answer

The value of $$\frac{\tan^2 \theta - \sin^2 \theta}{2 + \tan^2 \theta + \cot^2 \theta}$$ is:

A

$$\cosec^6 \theta$$

B

$$\cos^4 \theta$$

C

$$\sin^6 \theta$$

D

$$\sec^4 \theta$$

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Solution

$$\frac{\tan^2 \theta - \sin^2 \theta}{2 + \tan^2 \theta + \cot^2 \theta}$$

= $$\frac{\tan^2 \theta - \sin^2 \theta}{1 + \tan^2 \theta + 1 + \cot^2 \theta}$$

= $$\frac{\tan^2 \theta - \sin^2 \theta}{\sec^2 \theta + \cosec^2 \theta}$$

= $$\frac{\tan^2 \theta - \sin^2 \theta}{\frac{1}{\cos^2 \theta} +\frac{1}{\sin^2 \theta}}$$

= $$\frac{(\tan^2 \theta - \sin^2 \theta)(\sin^2\theta\cos^2 \theta)}{\cos^2 \theta + \sin^2 \theta}$$

= $$(\tan^2 \theta - \sin^2 \theta)(\sin^2\theta\cos^2 \theta)$$

= $$ \sin^4 \theta - \sin^4 \theta\cos^2 \theta$$

= $$ \sin^4 \theta(1 - \cos^2 \theta)$$ = $$ \sin^4 \theta \sin^2 \theta =  \sin^6 \theta$$

Question 72 Show answer

The value of the expression $$\cosec(85^\circ + \theta) - \sec(5^\circ - \theta) - \tan(55^\circ + \theta) + \cot(35^\circ - \theta)$$ is:

A

$$\frac{3}{2}$$

B

0

C

-1

D

1

View Explanation Hide Explanation

Solution

$$\cosec(85^\circ + \theta) - \sec(5^\circ - \theta) - \tan(55^\circ + \theta) + \cot(35^\circ - \theta)$$ 

=  $$\cosec(85^\circ + \theta) - \sec(90 - (85^\circ + \theta)) - \tan(55^\circ + \theta) + \cot(90 - (55^\circ + \theta))$$ 

= $$\cosec(85^\circ + \theta) - \cosec(85^\circ + \theta) - \tan(55^\circ + \theta) + \tan(55^\circ + \theta) $$  

= 0

Question 73 Show answer

If A lies in the first quadrant and $$6 \tan A = 5$$, then the value of $$\frac{8 \sin A - 4 \cos A}{\cos A + 2 \sin A}$$ is:

A

16

B

1

C

4

D

-2

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Solution

$$6 \tan A = 5$$

$$\tan A = \frac{5}{6}$$

$$\frac{8 \sin A - 4 \cos A}{\cos A + 2 \sin A}$$

= $$\frac{cos A(8 \tan A - 4)}{\cos A(1 + 2 \tan A)}$$

= $$\frac{8 \tan A - 4}{1 + 2 \tan A}$$

Put the value of $$\tan \theta$$,

= $$\frac{8 \times \frac{5}{6} - 4}{1 + 2 \times \frac{5}{6}}$$

= $$\frac{\frac{20}{3} - 4}{1 + \frac{5}{3}}$$

= $$\frac{\frac{8}{3}}{\frac{8}{3}}$$ = 1

Question 74 Show answer

If $$\sec \theta - \tan \theta = \frac{x}{y}, (0 < x < y)$$ and $$0^\circ < \theta < 90^\circ$$, then $$\sin \theta$$ is equal to:

A

$$\frac{x^2 + y^2}{y^2 - x^2}$$

B

$$\frac{x^2 + y^2}{2xy}$$

C

$$\frac{y^2 - x^2}{x^2 + y^2}$$

D

$$\frac{2xy}{x^2 + y^2}$$

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Solution

$$\sec \theta - \tan \theta = \frac{x}{y}$$ ---(1)

$$\frac{1}{\sec \theta - \tan \theta} = \frac{y}{x}$$

$$\frac{\sec^2 \theta - \tan^2 \theta}{\sec \theta - \tan \theta} = \frac{y}{x}$$
$$\frac{(\sec \theta - \tan \theta)(\sec \theta + \tan \theta)}{\sec \theta - \tan \theta} = \frac{y}{x}$$

$$\sec \theta + \tan \theta = \frac{y}{x}$$ ---(2)

From eq(1) and (2),

$$\sec \theta - \tan \theta + \sec \theta + \tan \theta =  \frac{x}{y} + \frac{y}{x}$$

$$2\sec \theta = \frac{x^2 + y^2}{xy} $$

$$\sec \theta = \frac{x^2 + y^2}{2xy} $$

$$\cos \theta = \frac{2xy}{x^2 + y^2} $$

Base = 2xy

Hypotenuses = {x^2 + y^2}

By Pythagoras,

$$(p)^2 + (2xy)^2 = (x^2 + y^2)^2$$

$$(p)^2 = x^4 + y^4 + 2x^2y^2 -  4x^2y^2

$$p = x^2 - y^2 = y^2 - x^2

sin$$\theta = p/h = \frac{y^2 - x^2}{x^2 + y^2}$$

Question 75 Show answer

Solve the following. $$\sin0^\circ \sin30^\circ \sin45^\circ \sin60^\circ \sin90^\circ = ?$$

A

0

B

4

C

$$\frac{\sqrt{6}}{8}$$

D

1

View Explanation Hide Explanation

Solution

$$\sin0^\circ \sin30^\circ \sin45^\circ \sin60^\circ \sin90^\circ = ?$$

$$Sin0^\circ = o$$

So,

? = 0

Question 76 Show answer

If $$5 \cos^{2} \theta + 1 = 3 \sin^{2} \theta, 0^\circ < \theta < 90^\circ$$, then what is the value of $$\frac{\tan \theta + \sec \theta}{\cot \theta + \cosec \theta}$$

A

$$\frac{3+2\sqrt{3}}{2}$$

B

$$\frac{2+3\sqrt{3}}{3}$$

C

$$\frac{2+3\sqrt{3}}{2}$$

D

$$\frac{3+2\sqrt{3}}{3}$$

View Explanation Hide Explanation

Solution

$$5 \cos^{2} \theta + 1 = 3 \sin^{2} \theta$$

$$5 \cos^{2} \theta + \sin^{2} \theta + \cos^{2} \theta = 3 \sin^{2} \theta$$

$$6 \cos^{2} \theta = 2 \sin^{2} \theta$$

$$\tan^{2} \theta = 3$$

$$\tan \theta = \sqrt3$$ 
$$\theta = 60\degree$$

$$\frac{\tan \theta + \sec \theta}{\cot \theta + \cosec \theta}$$

= $$\frac{\tan 60\degree + \sec 60\degree}{\cot 60\degree + \cosec 60\degree}$$

= $$\frac{\sqrt{3} + 2}{\frac{1}{\sqrt3} + \frac{2}{\sqrt3}}$$

= $$\frac{\sqrt{3} + 2}{\sqrt3}$$

= $$\frac{3 + 2\sqrt{3}}{3}$$

Question 77 Show answer

If $$6tan\theta-5\sqrt3sec\theta+12cot\theta=0,0^{0}<\theta<90^{0}$$, hen the value of $$(cosec\theta+sec\theta)$$ is:

A

$$\frac{3+2\sqrt{3}}{2}$$

B

$$\frac{2(3+2\sqrt{3})}{3}$$

C

$$\frac{2}{3}(3+\sqrt{3})$$

D

$$\frac{3+\sqrt{3}}{2}$$

View Explanation Hide Explanation

Solution

$$6tan\theta-5\sqrt3sec\theta+12cot\theta=0,0^{0}<\theta<90^{0}$$

$$\frac{6sin\theta}{cos\theta}-5\sqrt3(\frac{1}{cos\theta}) + \frac{12cos\theta}{sin\theta} = 0

$$\frac{6sin^2\theta - 5\sqrt3sin\theta + 12cos^2\theta}{sin\theta cos\theta}$$ = 0

$$\frac{6sin^2\theta + 6cos^2\theta- 5\sqrt3sin\theta + 6cos^2\theta}{sin\theta cos\theta}$$ = 0

$$\frac{6- 5\sqrt3sin\theta + 6cos^2\theta}{sin\theta cos\theta}$$ = 0

$$6- 5\sqrt3sin\theta + 6(1 - sin^2 \theta) = 0$$

$$6sin^2 \theta +  5\sqrt3sin\theta - 12 = 0$$

From LHS,

Put the $$\theta = 60 \degree$$

$$6sin^2 \theta + 5\sqrt3sin\theta - 12$$

$$6sin^2 60 \degree + 5\sqrt3sin60 \degree - 12$$

$$6(\frac{\sqrt3}{2})^2 + 5\sqrt3 \times \frac{\sqrt3}{2} - 12$$

$$\frac{18}{4} + \frac{15}{2} - 12$$ = 0

= RHS

$$(cosec\theta+sec\theta)$$

Put the $$\theta = 60 \degree$$,

= $$(cosec60 \degree+sec60 \degree)$$

= $$\frac{2}{\sqrt3} + 2$$

= $$\frac{2 + 2\sqrt3}{\sqrt3}$$

= $$\frac{2\sqrt3 + 6}{3}$$

= $$\frac{2}{3}(\sqrt3 + 3)$$

Question 78 Show answer

If $$\frac{\sin A + \cos A}{\cos A} = \frac{17}{12}$$, then the value of $$\frac{1 - \cos A}{\sin A}$$ is:

A

-5

B

1

C

$$\frac{5}{12}$$

D

$$\frac{1}{5}$$

View Explanation Hide Explanation

Solution

$$\frac{\sin A + \cos A}{\cos A} = \frac{17}{12}$$

$$\frac{\cos A(\frac{\sin A}{\cos A} + 1)}{\cos A} = \frac{17}{12}$$

tan A + 1 = $$\frac{17}{12}$$

tan A =  $$\frac{17}{12}$$ - 1 = $$\frac{5}{12}$$

By triplet 5-12-13,

Perpendicular = 5

base = 12

hypotenuse = 13 

so,

Sin A = perpendicular/hypotenuse = 5/13

cos A = base/hypotenuse = 12/13

Now,

$$\frac{1 - \cos A}{\sin A}$$

= $$\frac{1 - \frac{12}{13}}{\frac{5}{13}}$$ = 1/5

Question 79 Show answer

If $$5cot\theta=3$$, find the value of $$\frac{6sin\theta-3cos\theta}{7sin\theta+3cos\theta}$$.

A

$$\frac{11}{40}$$

B

$$\frac{20}{41}$$

C

$$\frac{21}{44}$$

D

$$\frac{44}{21}$$

View Explanation Hide Explanation

Solution

$$5cot\theta=3$$

$$cot\theta=3/5$$

$$\frac{6sin\theta-3cos\theta}{7sin\theta+3cos\theta}$$

= $$\frac{sin\theta(6-\frac{3cos\theta}{sin\theta})}{sin\theta(7 + \frac{3cos\theta}{sin\theta})}$$

= $$\frac{6 - 3cot\theta}{7 + 3cot\theta}$$

Put the value of cot$$\theta$$,

= $$\frac{6 - 3 \times \frac{3}{5}}{7 + 3\times \frac{3}{5}}$$

= $$\frac{6 -\frac{9}{5}}{7 + \frac{9}{5}}$$

=$$\frac{21}{44}$$

Question 80 Show answer

If $$7 \cos^2 \theta + 3 \sin^2 \theta = 6, 0^\circ < \theta < 90^\circ,$$ then the value of $$\frac{\cot^2 2\theta + \sec^2 2\theta}{\tan^2 2\theta - \sin^2 2\theta}$$ is:

A

$$\frac{52}{27}$$

B

$$\frac{26}{15}$$

C

$$\frac{28}{27}$$

D

$$\frac{49}{45}$$

View Explanation Hide Explanation

Solution

$$7 \cos^2 \theta + 3 \sin^2 \theta = 6$$

$$7(1 - \sin^2\theta) + 3 \sin^2 \theta = 6$$

$$7 - 7\sin^2\theta + 3 \sin^2 \theta = 6$$

$$4\sin^2\theta = 1$$

$$\sin\theta = 1/2$$ 

($$For 0^\circ < \theta < 90^\circ,$$)

$$\theta = 30\degree$$

$$\frac{\cot^2 2\theta + \sec^2 2\theta}{\tan^2 2\theta - \sin^2 2\theta}$$

$$\frac{\cot^2 2 \times 30\degree + \sec^2 2 \times 30\degree}{\tan^2 2 \times 30\degree - \sin^2 2 \times 30\degree}$$

$$\frac{\cot^2 60\degree + \sec^2 60\degree}{\tan^2 60\degree - \sin^2 60\degree}$$

$$\frac{(\frac{1}{\sqrt3})^2 + 2^2}{(\sqrt{3})^2 - (\frac{\sqrt3}{2})^2}$$

$$\frac{\frac{1}{3} + 4}{3 - \frac{3}{4}}$$

$$\frac{\frac{13}{3} }{\frac{9}{4}}$$ = 52/27

Question 81 Show answer

Seema flies a kite on a 16 m string at an inclination of $$60^{0}$$. What is the height (h) of the kite above the ground?

A

$$4\sqrt3$$ m

B

$$8\sqrt3$$ m

C

$$6\sqrt3$$ m

D

$$16\sqrt3$$ m

View Explanation Hide Explanation

Solution

sin$$\theta$$ = perpendicular/hypotenuse

 sin60$$\degree$$ = height/16

$$\frac{\sqrt3}{2}$$ = height/16

Height = 8$$\sqrt3$$

$$\therefore$$ Kite is 8$$\sqrt3$$ m above of the ground.

Question 82 Show answer

The value of $$\sqrt{\tan^2 60^\circ + \sin 90^\circ} - 2 \tan 45^\circ$$ is:

A

0

B

2

C

4

D

1

View Explanation Hide Explanation

Solution

$$\sqrt{\tan^2 60^\circ + \sin 90^\circ} - 2 \tan 45^\circ$$

On putting the values,

$$\sqrt{(\sqrt3)^2 + 1} - 2 \times 1$$

$$\sqrt{(4} - 2 $$ 

= 2 - 2 = 0

Question 83 Show answer

If A lies in third quadrant, and $$20 \tan A = 21$$, then the value of $$\frac{5 \sin A - 2 \cos A}{4 \cos A - \frac{5}{7} \sin A}$$

A

$$\frac{-65}{29}$$

B

$$\frac{1}{5}$$

C

$$\frac{5}{29}$$

D

$$\frac{13}{12}$$

View Explanation Hide Explanation

Solution

$$20 \tan A = 21$$

$$\tan A = 21/20$$

$$\frac{5 \sin A - 2 \cos A}{4 \cos A - \frac{5}{7} \sin A}$$

= $$\frac{\cos A(5 \frac{\sin A}{\cos A} - 2)}{\cos A(4 - \frac{5\sin A}{7\cos A})}$$

= $$\frac{5\tan A - 2}{4 - \frac{5}{7} \tan A}$$

On put the value of tan A,

= $$\frac{5 \times  \frac{21}{20}- 2}{4 - \frac{5}{7} \times \frac{21}{20}}$$

= $$\frac{\frac{13}{20}}{ \frac{13}{4}}$$ = $$\frac{1}{5}

Question 84 Show answer

The value of $$\frac{(\cos 9^\circ + \sin 81^\circ)(\sec 9^\circ + \cosec 81^\circ)}{2 \sin^2 63^\circ + 1 + 2 \sin^2 27^\circ}$$ is:

A

$$\frac{1}{2}$$

B

$$\frac{4}{3}$$

C

$$2$$

D

$$1$$

View Explanation Hide Explanation

Solution

$$\frac{(\cos 9^\circ + \sin 81^\circ)(\sec 9^\circ + \cosec 81^\circ)}{2 \sin^2 63^\circ + 1 + 2 \sin^2 27^\circ}$$

= $$\frac{(\cos(90 - 81) + \sin 81^\circ)(\sec (90 - 81) + \cosec 81^\circ)}{2 \sin^2 63^\circ + 1 + 2 \sin^2(90 - 63)}$$

= $$\frac{(\sin 81^\circ + \sin 81^\circ)(\cosec 81^\circ +\cosec 81^\circ)}{2 \sin^2 63^\circ + 1 + 2 \cos^2 63^\circ}$$

= $$\frac{(2\sin 81^\circ)(2\cosec 81^\circ}{2 + 1}$$ = $$\frac{4}{3}$$

Question 1 Show answer

A ladder leaning against a wall makes an angle $$\theta$$ with the horizontal ground such that $$\sin\theta = \frac{12}{13}$$. If the foot of the ladder is 7.5 m from the wall, then what is the height of the point where the top of the ladder touches the wall?

A

8 m

B

15 m

C

18 m

D

12 m

View Explanation Hide Explanation

Solution

Question 2 Show answer

$$\sin 18^\circ - \cos 72^\circ$$ is equal to:

A

$$\frac{1}{2}$$

B

0

C

2

D

1

Question 3 Show answer

A boy is standing near a pole which is 2.7 m high and the angle of elevation is 30“. The distance of the boy from the pole is ($$\surd3$$=1.73):

A

4.68 m

B

4.63 m

C

4.53 m

D

4.42 m

Question 4 Show answer

From the top of a 10 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of the foot of the tower is $$\phi$$ , such that tan $$\phi = \frac{2}{3}$$ . What is the height of the tower to nearest metres?

A

35 m

B

36 m

C

34 m

D

33 m

Question 5 Show answer

$$\frac{\cosec 31^\circ}{\sec 59^\circ}$$ is equal to:

A

1

B

3

C

0

D

2

Question 6 Show answer

$$(1+\cot^2\theta)(1-\cos^2\theta)$$ is equal to:

A

0

B

1

C

Not defined

D

$$\frac{1}{2}$$

View Explanation Hide Explanation

Solution

$$\left(1+\cot^2\theta\ \right)\left(1-\cos^2\theta\ \right)$$

$$\left(1+\frac{\cos^{^2}\theta\ }{\sin^2\theta\ }\right)\left(1-\cos^2\theta\ \right)$$

$$\left(1+\frac{\cos^{^2}\theta\ }{\sin^2\theta\ }\right)\left(\sin^2\theta\ \right)$$

$$\left(\sin^2\theta\ +\cos^2\theta\ \right)$$

$$1$$

Question 7 Show answer

From the top of a hill 96 m high. the angles of depression of two cars parked on the same side of the hill (at same level as the base of the hill) are $$30^\circ$$ and $$60^\circ$$ respectively. The distance between the cars is:
(use $$\surd3=1.73$$ and round off to nearest whole number.)

A

165 m

B

111 m

C

220 m

D

243 m

Question 8 Show answer

If $$\sec 2x = \cosec (3x - 45^\circ )$$. then $$x$$ is equal to:

A

$$27^\circ$$

B

$$40^\circ$$

C

$$45^\circ$$

D

$$35^\circ$$

Question 9 Show answer

The value of $$\frac{\sin^2 60^\circ + \cos^2 30^\circ - \sec 35^\circ.\sin 55^\circ}{\sec 60^\circ+\cosec 30^\circ}$$ is equal to:

A

$$\frac{1}{3}$$

B

$$\frac{1}{4}$$

C

$$\frac{1}{2}$$

D

$$\frac{1}{8}$$

Question 10 Show answer

The value of $$\sin^2 30^\circ .\cos^2 45^\circ + 2\tan^2 30^\circ - \sec^2 60^\circ $$ is equal to:

A

$$-\frac{13}{12}$$

B

$$-\frac{77}{24}$$

C

$$-\frac{25}{12}$$

D

$$-\frac{1}{12}$$

Question 11 Show answer

From the top of 75 m high tower, the angle of depression of two points P and Q on opposite side of the base of the tower on level ground is $$\theta$$ and $$\phi$$ , such that $$\tan \theta = \frac{3}{4}$$ and $$\tan \phi = \frac{5}{8}$$. What is the distance between the points P and Q?

A

190 m

B

200 m

C

180 m

D

220 m

Question 12 Show answer

From the top of a 120 m high tower, the angle of depression of the top of a pole is $$45^\circ$$ and the angle of depression of the foot of the pole is $$\theta$$, such that $$\tan \theta = \frac{3}{2}$$, What is the height of the pole?

A

80 m

B

40 m

C

60 m

D

75 m

Question 13 Show answer

From the top of a 12 m high building, the angle of elevation of the top of a tower is $$60^\circ$$ and the angle of depression of the foot of the tower is $$\theta$$ , such that $$\tan\theta=\frac{3}{4}$$ What is the height of the tower $$( \sqrt3=1.73)$$ ?

A

37.95 m

B

39.68 m

C

41.41 m

D

36.22 m

View Explanation Hide Explanation

Solution

Question 14 Show answer

If $$2\cos^2 \theta + 3\sin \theta = 3$$, where $$0^\circ < \theta < 90^\circ$$, then what is the value of $$\sin^2 2\theta + \cos^2 \theta + \tan^2 2\theta + \cosec^2 2\theta$$?

A

$$\frac{35}{12}$$

B

$$\frac{29}{3}$$

C

$$\frac{35}{6}$$

D

$$\frac{29}{6}$$

View Explanation Hide Explanation

Solution

$$2\cos^2 \theta + 3\sin \theta = 3$$
$$(2 - 2\sin^2 \theta) + 3\sin \theta - 3 = 0$$
$$2\sin^2 \theta - 3\sin \theta + 1 = 0$$
$$2\sin^2 \theta - 2\sin \theta - \sin \theta + 1 = 0$$
$$2\sin\theta(\sin\theta - 1) - 1(\sin\theta - 1) = 0$$
$$\sin\theta = 1 or \sin\theta = 1/2 $$ 
Here $$0^\circ < \theta < 90^\circ$$,
$$\sin\theta = 1/2$$
$$\theta = 30\degree$$
Now,
$$\sin^2 2\theta + \cos^2 \theta + \tan^2 2\theta + \cosec^2 2\theta$$
= $$\sin^2 2\times 30\degree + \cos^2 \times 30\degree + \tan^2 2\times 30\degree + \cosec^2 2\times 30\degree$$
= $$\sin^2 60\degree + \cos^2 30\degree + \tan^260\degree + \cosec^260\degree$$
= $$(\frac{\sqrt{3}}{2})^2 + (\frac{\sqrt{3}}{2})^2 + (\sqrt{3})^2 + (\frac{2}{\sqrt{3}})^2$$
=$$\frac{3}{4} + \frac{3}{4} + 3 + \frac{4}{3} = \frac{35}{6}$$

Question 15 Show answer

If $$ x = a  sin\theta - b  cos \theta, y = a cos \theta  + b sin \theta $$  , then which of the following is true?

A

$$ x^2  + y^2  = a^2  + b^2 $$

B

$$ \frac{x^2}{y^2}  + \frac{a^2}{b^2} = 1 $$

C

$$ x^2  + y^2  = a^2  - b^2 $$ 

D

$$ \frac{x^2}{a^2}  + \frac{y^2}{b^2} = 1 $$ 

View Explanation Hide Explanation

Solution

solution

$$ x = a  sin\theta - b  cos \theta$$  {squaring x}

 $$ y = a cos \theta  + b sin \theta $$ {squaring y}

$$ x^2 = a ^2 sin^2\theta + b^2  cos^2 \theta -2absin\theta cos \theta$$

$$ y^2 = a ^2 sin^2\theta + b^2  cos^2 \theta +2absin\theta cos \theta$$

adding both 

we get

$$ x^2 $$+ $$ y^2 $$   = $$a ^2 sin^2\theta + b^2  cos^2 \theta -2absin\theta cos \theta$$ + $$a ^2 sin^2\theta + b^2  cos^2 \theta +2absin\theta cos \theta$$

$$ x^2 $$+ $$ y^2 $$ = $$a ^2 sin^2\theta + b^2  cos^2 \theta + a ^2 sin^2\theta + b^2  cos^2 \theta$$    { $$\because cos^2 \theta +sin^2 \theta = 1$$}

$$ x^2 $$+ $$ y^2 $$ = $$a ^2 (sin^2\theta +  sin^2\theta) +  b^2  (cos^2 \theta +  cos^2 \theta)$$

$$ x^2 $$+ $$ y^2 $$ = $$ a^2 $$+ $$ b^2 $$

         

Question 16 Show answer

The shadow of a tower. when the angle of elevation of the sun is $$60^\circ$$ is found to be 15 in shorter than when it is $$45^\circ$$. The height of the tower is:

A

26.5 m

B

35.5 m

C

41.5 m

D

20.5 m

Question 17 Show answer

When the sun's angle of depression changes from $$30^\circ$$ to $$60^\circ$$. the length of the shadow of a tower decreases by 70 m. What is the height of the tower?

A

45.65 m

B

60.55 m

C

65.55 m

D

36.55 m

View Explanation Hide Explanation

Solution

Let the initial distance of the shadow from the foot of the tower be x m
Then, New distance will be (x-70) m 

$$tan 30^\circ = \dfrac{h}{x}$$

=> $$\dfrac{1}{\sqrt{3}} = \dfrac{h}{x}$$

=> $$x = h\sqrt{3}$$
$$tan 60^\circ = \dfrac{h}{x-70}$$

=> $$\sqrt{3} = \dfrac{h}{x-70}$$

=> $$\sqrt{3} = \dfrac{h}{h\sqrt{3}-70}$$

=> $$h = \sqrt{3}(h\sqrt{3}-70)$$
=> $$h = 3h - 70\sqrt{3}$$
=> $$2h = 70\sqrt{3}$$
=> $$h = 35\sqrt{3} = 35\times1.73 = 60.55 m$$
Therefore, The height of the tower = 60.55 m

Question 18 Show answer

The angle of elevation of a flying drone from a point on the ground is $$60^\circ$$. After flying for 5 seconds the angle of elevation drops to $$30^\circ$$. If the drone is flying horizontally at a constant height of $$1000\surd3$$ m, the distance travelled by the drone is:

A

2000 m

B

1000 m

C

3000 m

D

4000 m

Video Solution

Question 19 Show answer

$$(\cosec A - \sin A)^2 + (\sec A - \cos A)^2 - (\cot A - \tan A)^2$$ is equal to.

A

1

B

2

C

0

D

-1

View Explanation Hide Explanation

Solution

$$(\cosec A - \sin A)^2 + (\sec A - \cos A)^2 - (\cot A - \tan A)^2$$
$$= cosec^2A+sin^2A-2cosecAsinA + sec^2A+cos^2A-2secAcosA - (cot^2A+tan^2A-2cotAtanA)$$
$$= cosec^2A+sin^2A-2+sec^2A+cos^2A-2-cot^2A-tan^2A+2$$
$$= (cosec^2A-cot^2A)+(sin^2A+cos^2A)+(sec^2A-tan^2A)-2$$
$$= 3-2 = 1$$

Question 20 Show answer

What is the angle of elevation of the sun from the top of a vertical pole when its height is equal to the length of its shadow?

A

$$90^\circ$$

B

$$60^\circ$$

C

$$45^\circ$$

D

$$30^\circ$$

Question 21 Show answer

If $$x=a\ \cos\theta+b\ \sin\theta$$ and $$y=a\ \sin\theta-b\ \cos\theta$$, the value of $$x^2+y^2$$ is:

A

$$a^2-b^2$$

B

$$a-b$$

C

$$a^2+b^2$$

D

$$a+b$$

View Explanation Hide Explanation

Solution

$$x=aCos\theta + bSin\theta $$

$$x^2=(aCos\theta +bSin\theta)^2$$

$$=a^2Cos^2\theta+b^Sin^2\theta+2ab$$

$$Cos\theta Sin\theta $$.

Similarly,

$$y^2=b^2Cos^2\theta +a^2Sin^2\theta$$

$$-2abCos\theta Sin\theta$$.

So,$$x^2+y^2$$

$$=(a^2+b^2)Cos^2\theta + (a^2+b^2)$$

$$Sin^2\theta$$.

$$=(a^2+b^2)(Cos^2\theta+Sin^2\theta).$$

we know,$$Cos^2\theta +Sin^2\theta=1$$.

So,$$x^2+y^2=a^2+b^2.$$

Option C is correct.

Question 22 Show answer

A and B are standing on the same side of a wall and observe that the angles of elevation to the top of the wall are $$45^\circ$$ and $$60^\circ$$ respectively. If the height of the wall is 50 M. the distance between A and B is: (Use $$\sqrt3=1.73$$ and $$\sqrt2=1.41$$)

A

25.07 m

B

21.10 m

C

17.38 m

D

14.65 m

View Explanation Hide Explanation

Solution

So, $$50/(distance A)=tan45°=1$$

or, distance of $$A=50 m.$$

$$50/(distance of B)=tan60°=√3$$

or,distance of B $$=50/√3=28.86m.$$

So,distance between A and B 

$$=50-28.86=21.14.$$

So, Option B is correct.

Question 23 Show answer

The value of $$\left[ \frac{\sin^2 24^\circ+\sin^2 66^\circ}{\cos^2 24^\circ+\cos^2 66^\circ} + \sin^2 61^\circ + \cos 61^\circ \sin 29^\circ \right]$$ is equal to:

A

1

B

3

C

0

D

2

View Explanation Hide Explanation

Solution

we know that :
$$\sin^2\theta\ \ =\ \cos^2\left(90-\theta\ \right)$$
$$\sin\theta\ =\cos\left(90-\theta\ \right)$$
And $$\sin^2\theta\ +\cos^2\theta\ \ =1$$
Now using the above three identities
we get 1+sin^261+sin61cos29 = 1+$$1+\sin^261+\cos^261\ =2$$

Question 24 Show answer

If $$\tan x = \cot (45^\circ + 2x)$$, then what is the value of $$x$$?

A

$$45^\circ$$

B

$$20^\circ$$

C

$$15^\circ$$

D

$$\frac{45^\circ}{2}$$

Question 25 Show answer

The value of $$4\sin^230^\circ+3\cot^2 60^\circ$$ is:

A

$$1$$

B

$$\frac{1}{\sqrt{3}}$$

C

$$2$$

D

$$0$$

View Explanation Hide Explanation

Solution

$$Sin30°=1/2$$ and $$Cot60°=1/√3$$

So,$$4×Sin^230°=4×(1/4)=1$$

and $$3×Cot^260°=3×(1/3)=1$$

So,Sum of these two is 2.

Option C is correct.

Question 26 Show answer

The value of $$\cos^2 45^\circ + \sin^2 30^\circ - \sin^2 60^\circ$$ is equal to.

A

$$\frac{3}{2}$$

B

$$\frac{1}{2}$$

C

0

D

1

Question 27 Show answer

A ladder leaning against a wall makes an angle of 60° with the horizontal. If the foot of the ladder is 10 m away from the wall. what is the length of the ladder?

A

34.6 m

B

40 m

C

17.3 m

D

20 m

View Explanation Hide Explanation

Solution

Given, 

$$\angle\ ACB=\ 60^{\circ\ }$$

BC= 10 cm

We know that, $$\cos\ \theta\ =\frac{Base}{Hypotenuse}$$

$$\cos\ \theta\ =\frac{BC}{AC}$$

$$\cos\ 60^{\circ\ }=\ \frac{10}{AC}$$

$$\frac{1}{2}\ =\ \frac{10}{AC}$$

$$AC=20m$$

Question 28 Show answer

If $$2\sin 3\theta = 1$$, then the value of $$\theta$$ is:

A

$$30^\circ$$

B

$$20^\circ$$

C

$$10^\circ$$

D

$$45^\circ$$

Question 29 Show answer

The value of $$\frac{\sin30^\circ-\cos60^\circ+\cot^2 45^\circ}{\cos30^\circ-\tan 45^\circ+\sin 90^\circ}$$ is equal to:

A

$$\frac{\sqrt{3}}{4}$$

B

$$\frac{2\sqrt{3}}{3}$$

C

$$\frac{\sqrt{3}}{2}$$

D

$$\frac{3}{2}$$

View Explanation Hide Explanation

Solution

we know that,

sin30=$$\frac{1}{2}$$. cos60=$$\frac{1}{2}$$ , cot45=$$1$$, cos30=$$\ \frac{\sqrt{\ 3}}{2}$$, sin90= $$1$$, tan45=$$1$$, 

putting values into equation we get,

$$\frac{\left\{\frac{1}{2\ }-\frac{1}{2}+1^{2\ }\right\}}{\frac{\sqrt{\ 3}}{2}-1+1}\ =\frac{2\sqrt{\ }3}{3}\ ans$$

Question 30 Show answer

If $$\sin (A+ B) = \frac{\sqrt3}{2}$$ and $$\tan (A— B) = \frac{1}{\sqrt3}$$, then $$(2A + 3B)$$ is equal to:

A

$$120^\circ$$

B

$$135^\circ$$

C

$$130^\circ$$

D

$$125^\circ$$

Question 31 Show answer

If the height of a pole and the distance between the pole and a man standing nearby are equal, what would be the angle of elevation to the top of the pole?

A

$$60^\circ$$

B

$$90^\circ$$

C

$$30^\circ$$

D

$$45^\circ$$

View Explanation Hide Explanation

Solution

Here,

$$height distance(h)=base distance(b).$$

So,$$tan(angle)=(h/b)=1=tan (45°).$$

So, angle of elevation$$=45°.$$

Option D is correct.

Question 32 Show answer

The top of a broken tree touches the ground at an angle of 60° and at a distance of 45 m from the base of the tree. The total height of the tree is: (Use $$\sqrt{3}=1.73$$ and $$\sqrt{2}=1.41$$)

A

167.85 m

B

153.45 m

C

141.3 m

D

137.24 m

View Explanation Hide Explanation

Solution

$$tan60°=(height of non broken part/45).$$

So,height of non broken part$$=45√3=77.94$$

So,

$$77.94^2+45^2=height of broken part^2$$

or,$$height of broken part=√8100=90.$$

So, the height of tree$$=77.94+90=167.94.$$

A is correct choice.

Question 33 Show answer

If $$5\cos \theta - 12 \sin \theta = 0$$,the value of $$\frac{2\sin \theta + \cos \theta}{\cos \theta - \sin \theta}$$is:

A

$$1\frac{75}{119}$$

B

$$3\frac{1}{7}$$

C

$$2\frac{34}{35}$$

D

$$3\frac{2}{3}$$

Question 34 Show answer

If $$\tan3x=\cot(30^\circ+2x)$$, then what is the value of $$x$$ ?

A

$$18^\circ$$

B

$$12^\circ$$

C

$$10^\circ$$

D

$$15^\circ$$

Question 35 Show answer

$$\frac{4}{3} \tan^2 60^\circ + 3 \cos^2 30^\circ - 2 \sec^2 30^\circ - \frac{3}{4} \cot^2 60^\circ$$ is equal to:

A

$$\frac{8}{3}$$

B

$$\frac{5}{4}$$

C

$$\frac{7}{3}$$

D

$$\frac{10}{3}$$

View Explanation Hide Explanation

Solution

$$\frac{4}{3} \tan^2 60^\circ + 3 \cos^2 30^\circ - 2 \sec^2 30^\circ - \frac{3}{4} \cot^2 60^\circ$$

$$= \dfrac{4}{3}\times(\sqrt{3})^2 + 3 \times (\dfrac{\sqrt{3}}{2})^2 - 2\times(\dfrac{2}{\sqrt{3}})^2 - \dfrac{3}{4}\times(\dfrac{1}{\sqrt{3}})^2$$

$$= \dfrac{4}{3}\times3+3\times\dfrac{3}{4}-2\times\dfrac{4}{3}-\dfrac{3}{4}\times\dfrac{1}{3}$$

$$= 4+\dfrac{9}{4}-\dfrac{8}{3}-\dfrac{1}{4}$$

$$= \dfrac{10}{3}$$

Question 36 Show answer

The value of $$\cot^2 A - \frac{1}{\sin^2 A}$$ is equal to:

A

0

B

-1

C

-2

D

1

Question 37 Show answer

If $$4\tan\theta=3,\frac{5\sin\theta-3\cos\theta}{5\sin\theta+3\cos\theta}$$ is equal to:

A

$$\frac{1}{9}$$

B

$$\frac{1}{3}$$

C

$$9$$

D

$$3$$

View Explanation Hide Explanation

Solution

$$4\tan\theta\ =3.$$

or,$$\frac{\sin\ \theta\ \ }{\cos\ \theta\ }=\ \frac{\ 3}{4}.$$

Let say,$$\sin\theta\ =3k\ and\ \cos\theta\ =4k.$$

So,$$\ \frac{\ 5\sin\ \theta\ -3\cos\ \theta\ }{5\sin\ \theta\ +3\cos\ \theta\ }$$

$$\ =\frac{\ 5\times\ 3k\ -3\times\ 4k\ }{5\times\ 3k\ +3\times\ 4k\ }$$

$$\ =\frac{\ 15k\ -12k\ }{15k\ +12k\ }$$

$$\ =\frac{\ 3k\ }{27k\ }$$

$$\ =\frac{\ 1\ }{9}.$$

A is correct choice.

Question 38 Show answer

The length of shadow of a vertical pole on the ground is 24 m. If the angle of elevation of the sun at that time is $$\theta$$, such that $$\sin \theta = \frac{5}{13}$$ , then what is the height of the pole?

A

18 m

B

12 m

C

8 m

D

10 m

Question 39 Show answer

A girl 1.2 m tall can just see the sun over a 3.62 m tall wall which is 2.42 m away from her. The angle of elevation of the sun is:

A

$$60^\circ$$

B

$$30^\circ$$

C

$$90^\circ$$

D

$$45^\circ$$

Question 40 Show answer

From the top of 120 m high lighthouse, the angle of depression of two ships on opposite side of the base of the lighthouse is $$30^\circ$$ and $$60^\circ$$. What is the distance between the ships? (rounded oft)

A

327 m

B

177 m

C

277 m

D

127 m

Question 41 Show answer

From a point P on a level ground, the angle of elevation of the top of a tower is $$30^\circ$$. If the tower is 270 m high. the distance of point P from the foot of the tower is:

A

476.65 m

B

367.65 m

C

467.65 m

D

376.65 m

View Explanation Hide Explanation Video Solution

Solution

Given, Height of the tower = 270 m
Let the distance from the foot of the tower to P be x m

tan 30 = $$\dfrac{AB}{BP}$$

$$\dfrac{1}{\sqrt{3}} = \dfrac{270}{x}$$

$$x = 270\sqrt{3} = 270\times1.732 = 467.65 m$$

Question 42 Show answer

If $$\cos \theta = \frac{1}{\surd10}$$, then $$\tan \theta$$ is equal to:

A

3

B

$$\surd3$$

C

$$ \frac{1}{\surd3}$$

D

$$\frac{1}{3}$$

Question 43 Show answer

The angle of elevation of top of a tower from a point P, on the ground is $$\theta$$ such that tan $$\theta = \frac{12}{5}$$ . If distance of the point P, from the base of the tower is 75 m, what is the height of the tower?

A

160 m

B

200 m

C

190 m

D

180 m

Question 44 Show answer

$$1 + \frac{\tan^2 A}{1 + \sec A}$$ is equal to:

A

$$\sec A$$

B

$$\cosec A$$

C

$$\cos A$$

D

$$\sin A$$

Question 45 Show answer

The value of the expression $$(\cos^6 \theta + \sin^6 \theta - 1)(\tan^2 \theta + \cot^2 \theta + 2)$$ is:

A

0

B

-1

C

-3

D

1

View Explanation Hide Explanation

Solution

$$(\cos^6 \theta + \sin^6 \theta - 1)(\tan^2 \theta + \cot^2 \theta + 2)$$
Assume any value of $$\theta$$,
Let the $$\theta$$ be 45$$\degree$$,
= $$(\cos^6 45\degree + \sin^6 45\degree - 1)(\tan^2 45\degree + \cot^2 45\degree + 2)$$
= $$((\frac{1}{\sqrt{2}})^6 + (\frac{1}{\sqrt{2}})^6 - 1)(1 + 1 + 2)$$
= $$((\frac{1}{8}) + (\frac{1}{8}) - 1)(4)$$
= $$\frac{3}{4} \times 4$$
= -3

Question 46 Show answer

The value of $$\left(\frac{sinA}{1-cosA} + \frac{1-cosA}{sinA}\right) \div \left(\frac{cot^2A}{1+cosecA} + 1\right)$$ is:

A

$$\frac{3}{2}$$

B

$$\frac{1}{2}$$

C

1

D

2

View Explanation Hide Explanation

Solution

$$\left(\frac{sinA}{1-cosA} + \frac{1-cosA}{sinA}\right) \div \left(\frac{cot^2A}{1+cosecA} + 1\right)$$
Let the value of $$\theta = 45\degree$$,
$$\left(\frac{\frac{1}{\sqrt{2}}}{1- \frac{1}{\sqrt{2}}} + \frac{1- \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}\right) \div \left(\frac{1}{1+\sqrt{2}} + 1\right)$$

=$$\left(\frac{\frac{1}{2} + (1- \frac{1}{\sqrt{2}})^2}{(\frac{1}{\sqrt{2}})(1- \frac{1}{\sqrt{2}})}\right) \div \left(\frac{1 + 1+\sqrt{2}}{1+\sqrt{2}}\right)$$

= $$\left(\frac{\frac{1}{2} + 1 + \frac{1}{2} - \sqrt{2}}{(\frac{1}{\sqrt{2}})(1- \frac{1}{\sqrt{2}})}\right) \div \left(\frac{2+\sqrt{2}}{1+\sqrt{2}}\right)$$

= $$\left(\frac{2 - \sqrt{2}}{(\frac{1}{\sqrt{2}}- \frac{1}{2})}\right) \div \left(\frac{2+\sqrt{2}}{1+\sqrt{2}}\right)$$

= $$\left(\frac{2 - \sqrt{2}}{\frac{2 - \sqrt{2}}{2\sqrt{2}}}\right) \div \left(\frac{2+\sqrt{2}}{1+\sqrt{2}}\right)$$

= $$2\sqrt{2} \times \frac{1+\sqrt{2}}{2+\sqrt{2}}$$ = 2

Question 47 Show answer

The value of $$(1 + \cot \theta - \cosec \theta )(1 + \cos \theta + \sin \theta)\sec \theta = ?$$

A

-2

B

2

C

$$\sec \theta \cosec \theta$$

D

$$\sin \theta \cos \theta$$

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Solution

$$(1 + \cot \theta - \cosec \theta )(1 + \cos \theta + \sin \theta)\sec \theta$$
Let $$\theta = 45\degree$$,
$$(1 + \cot 45\degree - \cosec 45\degree )(1 + \cos 45\degree + \sin 45\degree)\sec 45\degree$$
= $$(1 + 1 - \sqrt{2})(1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})\sqrt{2}$$
= $$(2 - \sqrt{2})(\frac{2 + \sqrt{2}}{\sqrt{2}})\sqrt{2}$$
= $$(2 - \sqrt{2})(2 + \sqrt{2})$$
= $$(2^2 - (\sqrt{2})^2)$$ = 4 - 2 = 2

Question 48 Show answer

$$\frac{(2 \sin A)(1 + \sin A)}{1 + \sin A + \cos A}$$ is equal to:

A

$$1 + \sin A - \cos A$$

B

$$1 - \sin A \cos A$$

C

$$1 + \cos A - \sin A$$

D

$$1 + \sin A \cos A$$

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Solution

$$\frac{(2 \sin A)(1 + \sin A)}{1 + \sin A + \cos A}$$
= $$\frac{(2 \sin A + 2\sin^2 A)}{1 + \sin A + \cos A}$$
= $$\frac{(2 \sin A + 2 - 2\cos^2 A)}{1 + \sin A + \cos A}$$
($$\because \sin^2 A + \cos^2 A = 1$$)
= $$\frac{(2 \sin A + 1 + \sin^2 A + \cos^2 A - 2\cos^2 A)}{1 + \sin A + \cos A}$$
= $$\frac{((\sin A + 1)^2 - \cos^2 A)}{1 + \sin A + \cos A}$$
($$\because (a)^2 - (b)^2 = (a + b)(a - b)$$)
= $$\frac{(1 + \sin A + \cos A)(1 + \sin A - \cos A)}{1 + \sin A + \cos A}$$
= $$(1 + \sin A - \cos A)$$

Question 49 Show answer

If $$\frac{1 + sin \phi}{1 - sin \phi} = \frac{p^2}{q^2}$$, then $$sec \phi$$ is equal to 

A

$$\frac{2 P^2 Q^2}{p^2 + q^2}$$

B

$$\frac{1}{2} \left(\frac{q}{p} + \frac{p}{q}\right)$$

C

$$\frac{1}{p^2}+\frac{1}{q^2}$$

D

$$\frac{ P^2 q^2}{p^2+q^2}$$

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Solution

$$\frac{1 + sin \phi}{1 - sin \phi} = \frac{p^2}{q^2}$$
By componendo dividendo rule,
$$\frac{(1 + sin \phi) + (1 - sin \phi)}{(1 + sin \phi) - (1 - sin \phi)} = \frac{p^2 + q^2}{p^2 - q^2}$$
$$\frac{2}{2 sin \phi} = \frac{p^2 + q^2}{p^2 - q^2}$$
$$ sin \phi = \frac{p^2 - q^2}{p^2 + q^2}$$ = $$\frac{perpendicular}{hypotenuse}$$
By the Pythagoras theorem,
Base = $$\sqrt{(p^2 + q^2)^2- (p^2 - q^2)^2 } = \sqrt{4p^2q^2} = 2pq$$
$$sec \phi$$ =  $$\frac{hypotenuse}{base} = \frac{p^2 + q^2}{2pq}$$

Question 50 Show answer

If $$ 7sin^2 \theta  + 3cos^2\theta  = 4 $$, then the value of $$ tan \theta is ( \theta $$ is acute)

A

$$ \frac{1}{\sqrt{3}} $$

B

$$ \frac{1}{\sqrt{2}} $$

C

$$ 1 $$

D

$$ {\sqrt{3}} $$

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Solution

Given that $$ 7sin^2 \theta + 3cos^2\theta = 4 $$

=> $$ 3sin^2 \theta + 3cos^2\theta + 4 sin^2 \theta = 4 $$

=> $$ 3(sin^2 \theta + cos^2\theta) + 4 sin^2 \theta = 4 $$

=> $$ 3 + 4 sin^2 \theta = 4 $$

=> $$ 4 sin^2 \theta = \frac{4}{3} $$

=> $$ sin^2 \theta = \frac{1}{3} $$

=> $$ sin \theta = \frac{1}{\sqrt{3}} $$

=> $$ \theta $$ = $$30^\circ $$

Therefore, $$ tan30^\circ $$ = $$ \frac{1}{\sqrt{3}}$$

Question 51 Show answer

If $$ \tan A = n \tan B  and   \sin A = m \sin B,$$  then the value of $$ \cos^2 A$$ is

A

$$ \frac{m^2 +1}{n^2 +1} $$

B

$$ \frac{m^2 -1}{n^2 -1} $$

C

$$ \frac{m^2 +1}{n^2 -1} $$

D

$$ \frac{m^2 - 1}{n^2 +1} $$

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Solution

Given that $$ \tan A = n \tan B $$ and  $$\sin A = m \sin B$$ ---- (1)

=> $$\frac{\sin A}{\cos A}$$ = n$$\frac{\sin B}{\cos B}$$

=> $$\frac{m\sin B}{\cos A}$$ = n $$\frac{\sin B}{\cos B}$$

=> $$\frac{\cos A}{\cos B}$$ = $$\frac {m}{n}$$ ----- (2)

Squaring equation (1), we get

=> $$\sin^2 A = m^2 \sin^2 B$$

=> $$ 1-\cos^2 A = m^2 (1-cos ^2 B) $$

=> $$ cos ^2 B = \frac {m^2 -1 + \cos^2 A}{m^2}$$ ---- (3)

Squaring equation (2) and substituting equation (3) in equation (2), we get

=> $$ cos^2 A = [\frac{m^2}{n^2}][\frac {m^2 -1 + \cos^2 A}{m^2}]$$

=> $$ n^2 cos^2 A= m^2 -1 + \cos^2 A $$

=> $$ cos ^2 A = \frac {m^2 -1}{n^2 -1} $$

Question 52 Show answer

The value of $$\frac{ sec\phi\left(1-\sin\phi\right)\left(\sin \phi +\cos \phi \right)\left(\sec \phi +\tan \phi \right)}{ \sin\phi\left(1+\tan\phi\right)+\cos\phi\left(1+\cot\phi\right)}$$ is equal to:

A

$$2cos\phi$$

B

$$cosec\phi sec\phi$$

C

$$2sin\phi$$

D

$$sin\phi cos\phi$$

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Solution

$$\frac{ sec\phi\left(1-\sin\phi\right)\left(\sin \phi +\cos \phi \right)\left(\sec \phi +\tan \phi \right)}{ \sin\phi\left(1+\tan\phi\right)+\cos\phi\left(1+\cot\phi\right)}$$
Put the value of $$\phi = 60\degree$$,
$$\frac{ sec60\degree\left(1-\sin60\degree\right)\left(\sin60\degree +\cos 60\degree \right)\left(\sec 60\degree +\tan 60\degree \right)}{ \sin60\degree\left(1+\tan60\degree\right)+\cos60\degree\left(1+\cot60\degree\right)}$$
= $$\frac{ 2\left(1-\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2} +\frac{1}{2} \right)\left(2 +\sqrt{3} \right)}{ \frac{\sqrt{3}}{2}\left(1+\sqrt{3}\right)+\frac{1}{2}\left(1+\frac{1}{\sqrt{3}}\right)}$$
= $$\frac{\left(2-\sqrt{3}\right)\left(\frac{\sqrt{3} + 1}{2} \right)\left(2 +\sqrt{3} \right)}{ \frac{\sqrt{3}}{2}\left(1 + \sqrt{3}\right)+\frac{1}{2}\left(\frac{1 + \sqrt{3}}{\sqrt{3}}\right)}$$
= $$\frac{\left(2-\sqrt{3}\right)\left({\sqrt{3} + 1} \right)\left(2 +\sqrt{3} \right)}{ \sqrt{3}\left(1 + \sqrt{3}\right)+\left(\frac{1 + \sqrt{3}}{\sqrt{3}}\right)}$$

= $$\frac{\sqrt{3} + 1}{ \sqrt{3}\left(1 + \sqrt{3}\right)+\left(\frac{1 + \sqrt{3}}{\sqrt{3}}\right)}$$

= $$\frac{1}{ \sqrt{3}+\frac{1}{\sqrt{3}}}$$ = $$\frac{\sqrt{3}}{4}$$
From the option D,
$$sin\phi cos\phi$$
Put the value of $$\phi = 60\degree$$,
= $$sin60\degree cos60\degree$$
= $$\frac{\sqrt{3}}{2} \times \frac{1}{2}$$

= $$\frac{\sqrt{3}}{4}$$

Question 53 Show answer

If \sin A + \sin^A = 1,$$ then the value of $$ \cos^A + \cos^4 A is

A

$$ 1\frac{2}{3} $$

B

$$ 2 $$ 

C

$$ 1\frac{1}{2} $$

D

$$ 1 $$ 

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Solution

Information provided in the question is not in a understandable format. Please review and provide correct data.

Question 54 Show answer

The value of $$\frac{\sec^2 \theta}{\cosec^2 \theta} + \frac{\cosec^2 \theta}{\sec^2 \theta} - (\sec^2 \theta + \cosec^2 \theta)$$ is:

A

0

B

-2

C

2

D

1

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Solution

$$\frac{\sec^2 \theta}{\cosec^2 \theta} + \frac{\cosec^2 \theta}{\sec^2 \theta} - (\sec^2 \theta + \cosec^2 \theta)$$
Put the $$\theta = 60\degree$$
= $$\frac{\sec^2 60\degree}{\cosec^2 60\degree} + \frac{\cosec^2 60\degree}{\sec^2 60\degree} - (\sec^2 60\degree + \cosec^2 60\degree)$$
= $$\frac{4}{\frac{4}{3}} + \frac{\frac{4}{3}}{4} - (4 + \frac{4}{3})$$
= 3 + $$\frac{4}{12} - \frac{16}{3}$$
= $$\frac{-24}{12}$$ = -2

Question 55 Show answer

If $$\sin \theta = \sqrt{3} \cos \theta, 0^\circ < \theta < 90^\circ$$, then the value of $$2 \sin^2 \theta + \sec^2 \theta + \sin \theta \sec \theta + \cosec \theta$$ is:

A

$$\frac{33 + 10\sqrt{3}}{6}$$

B

$$\frac{19 + 10\sqrt{3}}{6}$$

C

$$\frac{33 + 10\sqrt{3}}{3}$$

D

$$\frac{19 + 10\sqrt{3}}{3}$$

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Solution

$$\sin \theta = \sqrt{3} \cos \theta, 0^\circ < \theta < 90^\circ$$
$$\Rightarrow  \frac{\sin \theta}{\cos \theta} = \sqrt{3}$$
$$\Rightarrow \tan \theta = \sqrt{3}$$ 
$$\Rightarrow \theta = 60\degree$$
Now,
$$2 \sin^2 \theta + \sec^2 \theta + \sin \theta \sec \theta + \cosec \theta$$
= $$2 \sin^2 60\degree + \sec^2 60\degree + \tan 60\degree + \cosec60\degree$$
= $$2 \times (\frac{\sqrt{3}}{2})^2 + 2^2 + \sqrt{3} + \frac{2}{\sqrt{3}}$$
= $$\frac{3}{2} + 4 + \sqrt{3} + \frac{2}{\sqrt{3}}$$
= $$\frac{3\sqrt{3} + 8\sqrt{3} + 6 + 4}{2\sqrt{3}}$$
= $$\frac{11\sqrt{3} + 10}{2\sqrt{3}}$$
= $$\frac{33 + 10\sqrt{3}}{6}$$
$$\therefore$$ The correct answer is option A.

Question 56 Show answer

If $$ \sec \theta - \tan \theta = \frac{1}{\sqrt{3}} $$ then value of $$ \sec\theta \tan \theta$$ is

A

$$ \frac{2}{{3}} $$

B

$$ \frac{2}{\sqrt{3}} $$

C

$$ \frac{1}{\sqrt{3}} $$

D

$$ \frac{4}{\sqrt{3}} $$

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Solution

Given that  $$ \sec \theta - \tan \theta = \frac{1}{\sqrt{3}} $$  ------> (1)

We know that $$sec^2 \theta - tan^2 \theta $$ = 1

=> $$(sec \theta - tan \theta)(sec \theta + tan \theta)$$ =1

=> $$sec \theta + tan \theta $$ = $$\sqrt{3}$$ ---------> (2)

Solving equations (1) and (2) , we get 

$$ sec \theta $$ = $$\frac{2}{ \sqrt{3}}$$ and $$ tan \theta $$ = $$\frac{1}{ \sqrt{3}}$$

Therefore $$ sec \theta tan \theta $$ = $$ \frac {2}{3} $$

Question 57 Show answer

The value of $$tan^2 \phi+cot^2 \phi-sec^2 \phi cosec^2 \phi$$ is equal to

A

-2

B

1

C

0

D

-1

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Solution

$$tan^2 \phi+cot^2 \phi-sec^2 \phi cosec^2 \phi$$
Put the $$\theta = 45\degree$$
= $$tan^2 45\degree + cot^2 45\degree - sec^2 45\degree cosec^2 45\degree$$
= $$1 + 1 - 2 \times 2$$
= 1 + 1 - 4 = -2

Question 58 Show answer

The value of $$\frac{(\cos 9^\circ + \sin 81^\circ)(\sec 9^\circ + \cosec 81^\circ)}{\sin 56^\circ sec 34^\circ + \cos 25^\circ \cosec 65^\circ}$$ is:

A

4

B

$$\frac{1}{2}$$

C

2

D

$$\frac{1}{4}$$

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Solution

$$\frac{(\cos 9^\circ + \sin 81^\circ)(\sec 9^\circ + \cosec 81^\circ)}{\sin 56^\circ sec 34^\circ + \cos 25^\circ \cosec 65^\circ}$$
= $$\frac{(\cos 9^\circ + \sin(90 - 9))(\sec 9^\circ + \cosec(90 - 9))}{\sin 56^\circ sec(90 - 56) + \cos 25^\circ \cosec(90 - 25)}$$
= $$\frac{(\cos 9^\circ + \cos 9^\circ)(\sec 9^\circ + \sec 9^\circ)}{\sin 56^\circ cosec 56^\circ + \cos 25^\circ \sec 25^\circ}$$
= $$\frac{(2\cos 9^\circ)(2 \sec 9^\circ)}{1 + 1}$$
= $$\frac{4}{2}$$ = 2

Question 59 Show answer

$$\left(\sec\phi-\tan\phi\right)^2\left(1+\sin\phi\right)^2\div sin^2 \phi$$ = ?

A

$$cos \phi$$

B

$$cot^2  \phi$$

C

$$sec \phi$$

D

$$cos^2 \phi$$

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Solution

$$\left(\sec\phi-\tan\phi\right)^2\left(1+\sin\phi\right)^2\div sin^2 \phi$$
Put the $$\theta = 45\degree$$,
= $$\left(\sqrt{2}-1\right)^2\left(1+\frac{1}{\sqrt{2}}\right)^2\div \frac{1}{{2}}$$
= $$\left(\sqrt{2}-1\right)^2\left(\frac{\sqrt{2} + 1}{\sqrt{2}}\right)^2\div \frac{1}{{2}}$$
= $$\left(\sqrt{2}-1\right)^2\left(\sqrt{2} + 1\right)^2$$
= $$(2 + 1 - 2\sqrt{2})(2 + 1 + 2\sqrt{2})$$
= $$(3 - 2\sqrt{2})(3 + 2\sqrt{2})$$
= 9 - 8 = 1
From the option B,
$$\cot^2 \phi$$
Put the $$\theta = 45\degree$$,
= $$\cot^2 45\degree$$
= 1

Question 60 Show answer

If $$\theta$$ lies in the first quadrant and $$\cos^2 \theta - \sin^2 \theta = \frac{1}{2}$$, then the value of $$\tan^2 2\theta + \sin^2 3\theta$$ is:

A

$$\frac{7}{2}$$

B

3

C

4

D

$$\frac{4}{3}$$

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Solution

$$\cos^2 \theta - \sin^2 \theta = \frac{1}{2}$$
$$\cos^2 \theta - (1 - \cos^2 \theta) = \frac{1}{2}$$
$$2\cos^2 \theta - 1 = \frac{1}{2}$$
$$2\cos^2 \theta = \frac{3}{2}$$
$$\cos^2 \theta = \frac{3}{4}$$
$$\cos \theta = \frac{\sqrt{3}}{2}$$
$$\theta = 30\degree$$
Now,
$$\tan^2 2\theta + \sin^2 3\theta$$
= $$\tan^2 2\times30\degree + \sin^2 3\times30\degree$$
= $$\tan^2 60\degree + \sin^2 90\degree$$
= 3 + 1 = 4

Question 61 Show answer

What is the value of $$\cosec(65^\circ + \theta) - \sec(25^\circ - \theta) + \tan^2 20^\circ - \cosec^2 70^\circ $$ ?

A

0

B

1

C

2

D

-1

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Solution

$$\cosec(65^\circ + \theta) - \sec(25^\circ - \theta) + \tan^2 20^\circ - \cosec^2 70^\circ $$
=$$\cosec(65^\circ + \theta) - \sec(90 - (65^\circ + \theta)) + \tan^2 20^\circ - \cosec^2 (90 - 20)$$
=$$\cosec(65^\circ + \theta) - \cosec(65^\circ + \theta) + \tan^2 20^\circ - \sec^2 20^\circ$$
=$$ \tan^2 20^\circ - \sec^2 20^\circ$$
=$$ -(\sec^2 20^\circ -\tan^2 20^\circ )$$ = -1

Question 62 Show answer

A businessman's earning increase by 25%in one year but decreases by 4%in the next. Goingby this pattern, after 5 years, his total earnings would be Rs.72000. Whatis his present earning?

A

Rs.10000

B

Rs.80000

C

Rs.40000

D

Rs.54000

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Solution

using the chaining method 

we can write 25% = $$\frac{1}{4}$$ , 4% = $$\frac{1}{25}$$

1st years increase 4-----> 5

2nd year decrease 25 ----> 24

3rd year increase 4--------> 5

4th year decrease 25 ----> 24

5th years increase 4-----> 5

-------------------------------------

intial --->final ratio  is  5--------> 9

after 5 years earning is 9 --->72000

present earning = 5 $$\times 8000$$ = rs 40000

Question 63 Show answer

$$\frac{\left(1+\cos\theta\right)^2+\sin^2\theta}{\left(\text{coec}^2\theta-1\right)\sin^2\theta}=$$

A

$$\cos\theta\left(1+\sin\theta\right)$$

B

$$2 \cos\theta\left(1+\sec\theta\right)$$

C

$$\sec\theta\left(1+\sin\theta\right)$$

D

$$2 \sec\theta\left(1+\sec\theta\right)$$

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Solution

$$\frac{\left(1+\cos\theta\right)^2+\sin^2\theta}{\left(\text{coec}^2\theta-1\right)\sin^2\theta}$$
=$$\frac{1 + \cos^2\theta + 2\cos\theta +\sin^2\theta}{1 -\sin^2\theta}$$
=$$\frac{2 + 2\cos\theta }{\cos^2\theta}$$
=$$2(\sec^2\theta+ \sec\theta)$$
=$$2\sec\theta(1 + \sec\theta)$$

Question 64 Show answer

The value of

 $$ cot41^\circ. cot42^\circ. cot43^\circ.co44. cot45^\circ.  cot46^\circ.cot47^\circ.cot48^\circ. cot49^\circ $$

A

0

B

1

C

$$\frac{\sqrt{3}}{2}$$

D

$$\frac{1}{\sqrt{2}}$$

Question 65 Show answer

The value of $$\frac{2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta)}{\cos^4 \theta - \sin^4 \theta - 2 \cos^2 \theta}$$ is:

A

-1

B

-2

C

2

D

1

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Solution

$$\frac{2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta)}{\cos^4 \theta - \sin^4 \theta - 2 \cos^2 \theta}$$
Put the $$\theta = 90\degree$$,
$$\frac{2(\sin^6 90\degree + \cos^690\degree) - 3(\sin^4 90\degree + \cos^4 90\degree)}{\cos^4 90\degree - \sin^4 90\degree - 2 \cos^2 90\degree}$$
= $$\frac{2(1 + 1) - 3(1 + 1)}{1 - 1 - 2 }$$
= $$\frac{4 - 6}{-2} = 1$$

Question 66 Show answer

What is the value of $$2 \sin 15$$° $$\cos 15$$° - $$4 \sin^3 15$$° $$\cos 15$$°?

A

$$\frac{3}{\surd3}$$

B

$$\frac{\surd3}{2}$$

C

$$\frac{\surd3}{4}$$

D

$$\frac{1}{2}$$

Question 67 Show answer

What is the value of $$[(\cos 7A + \cos 5A)÷(\sin 7A - \sin 5A)]$$?

A

$$\tan A$$

B

$$\tan 4A$$

C

$$\cot 4A$$

D

$$\cot A$$

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Solution

We have cos 7A +Cos 5A = 2cos 6A cos A   (1)  cos C +cos D = 2cos(C+D)/2 cos (C-D)/2
Now 
sin7A -sin5A = 2cos6AsinA       (2)                     sinC -sinD = 2cos(C+D)/2 sin(C-D)/2
Dividing (1) and (2)
we get $$\frac{\cos7A+\cos5A}{\sin7A-\sin5A}=\frac{\cos A}{\sin A}=\ \cot A$$

Question 68 Show answer

What is the value of $$\frac{2(1 - \sin^2 \theta)\cosec^2 \theta}{\cot^2 \theta(1 + \tan^2 \theta)} - 1$$?

A

$$\sin 2\theta$$

B

$$\sin^2 \theta$$

C

$$\cos^2 \theta$$

D

$$\cos 2\theta$$

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Solution

$$\frac{2(1-\sin^2\theta)\operatorname{cosec}^2\theta}{\cot^2\theta(1+\tan^2\theta)}-1\ .$$

$$=\frac{2\cos^2\theta\ \operatorname{cosec}^2\theta}{\frac{\cos^2\theta}{\sin^2\theta\ }.\sec^2\theta\ }-1\ .$$

$$=\frac{2\cos^2\theta\ \operatorname{cosec}^2\theta\sin^2\theta\ }{\cos^2\theta.\sec^2\theta\ }-1\ .$$

$$=\frac{2\ .\ 1.\ \ 1}{\sec^2\theta\ }-1\ .$$

$$=2\ \cos^2\theta\ -1\ .$$

$$=\cos2\theta\ \ .$$

D is correct choice.

Question 69 Show answer

What is the value of $$\frac{\left\{(\sin 4x + \sin 4y) [(\tan 2x - 2y)]\right\}}{(\sin 4x - \sin 4y)}$$?

A

$$\tan 2 (2x + 2y)$$

B

$$\tan^2$$

C

$$\cot (x - y)$$

D

$$\tan (2x + 2y)$$

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Solution

$$\frac{\left\{(\sin4x+\sin4y)[(\tan2x-2y)]\right\}}{(\sin4x-\sin4y)}$$

$$=\frac{2\times\sin\left(\frac{4x+4y}{2}\right)\times\ \cos\left(\frac{4x-4y}{2}\right)\left(\tan2x-2y\right)}{2\times\ \cos\left(\frac{4x+4y}{2}\right)\times\sin\left(\frac{4x-4y}{2}\right)}\ .$$

$$=\tan\left(2x+2y\right).\cot\left(2x-2y\right).\tan\left(2x-2y\right)\ .$$

$$=\tan\left(2x+2y\right)\ .$$

D is correct choice.

Question 70 Show answer

What is the value of $$\frac{\left[\sin (y - z) + \sin (y + z) + 2 \sin y\right]}{\left[\sin (x - z) + \sin (x + z) + 2 \sin x\right]}$$?

A

$$\cos x \sin y$$

B

$$\frac{(\sin y)}{(\sin x)}$$

C

$$\sin z$$

D

$$\sin x \tan y$$

Question 71 Show answer

What is the value of $$\frac{[(\sin x + \sin y) (\sin x - \sin y)]}{[(\cos x + \cos y) (\cos y - \cos x)]}?$$

A

0

B

1

C

-1

D

2

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Solution

$$\dfrac{[(\sin x + \sin y) (\sin x - \sin y)]}{[(\cos x + \cos y) (\cos y - \cos x)]}$$
=$$\dfrac{sin^{2}x-sin^{2}y}{cosxcosy-cosxcosy+cos^{2}x-cos^{2}y}$$
=$$\dfrac{sin^{2}x-sin^{2}y}{1-sin^{2}x-(1-sin^{2}y)}$$
=$$\dfrac{sin^{2}x-sin^{2}y}{sin^{2}x-sin^{2}y}$$
=1

Question 72 Show answer

What is the value of $$[(\sin 7x - \sin 5x) \div (\cos 7x + \cos 5x)] - [(\cos 6x - \cos 4x) \div (\sin 6x + \sin 4x)]$$?

A

1

B

$$2 \tan x$$

C

$$\tan 2 x$$

D

$$\tan (\frac{3x}{2})$$

Question 73 Show answer

If $$\sin x = \frac{1}{2}$$ and $$\sin y = \frac{2}{3}$$, then what is the value of $$\left[\frac{(6 \cos^2 x - 4 \cos^4 x)}{(18 \cos^2 y - 27 \cos^4 y)}\right]$$

A

$$\frac{27}{20}$$

B

$$\frac{15}{14}$$

C

$$\frac{25}{21}$$

D

$$\frac{17}{14}$$

Question 74 Show answer

$$\left(\frac{1-\tan\theta}{1-\cot\theta}\right)^2+1=$$

A

$$\cosec^2\theta$$

B

$$\sec^2\theta$$

C

$$\sin^2\theta$$

D

$$\cos^2\theta$$

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Solution

$$\left(\frac{1-\tan\theta}{1-\cot\theta}\right)^2+1$$
$$(\frac{1-\tan\theta}{1-\frac{1}{tan\theta}})^2+1$$
$$(\frac{1-\tan\theta}{\frac{tan\theta - 1}{tan\theta}})^2+1$$
$$(-tan\theta)^2 + 1$$
$$tan^2\theta + 1 = sec^2\theta$$

Question 75 Show answer

What is the value of $$[(\cos^3 2\theta + 3 \cos 2\theta) \div (\cos^6 \theta - \sin^6 \theta)]$$?

A

0

B

1

C

4

D

2

Question 76 Show answer

What is the value of $$\frac{[1 - \sin (90 - 2A)]}{[1 + \sin (90 + 2A)]}$$?

A

$$\sin A \cos A$$

B

$$\cot^2A$$

C

$$\tan^2A$$

D

$$\sin^2A \cos A$$

Question 77 Show answer

What is the value of $$\frac{(32 \cos^6 x - 48 \cos^4 x + 18 \cos^2 x - 1)}{[4 \sin x \cos x \sin (60 - x) \cos (60 - x) \sin (60 + x) \cos (60 + x)]}$$?

A

$$4 \tan 6x$$

B

$$4 \cot 6x$$

C

$$8 \cot 6x$$

D

$$8 \tan 6x$$

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Solution

We know :

$$\cos\ \left(2X\right)=2\cos^2X-1\ .$$

Now replace X=3x :

$$\cos6X=2\cos^23X-1\ .$$

Again, we know : $$\cos3X=4\cos^3X-3\cos X\ .$$

So, $$\cos6X=2\left(4\cos^3X-3\cos X\ \right)^2-1\ .$$

or, $$\cos6X=2\left(16\cos^6X+9\cos^2X\ -24\cos^3X\ \cos X\right)-1\ .$$

or, $$\cos6X=32\cos^6X+18\cos^2X\ -48\cos^4X\ -1\ .$$

Now, 

$$4 sin (60-x).sin x. sin(60+x)$$

$$= 4 (sin 60.cos x - cos60 sin x).sin x. (sin 60.cos x + cos60 sin x)$$

$$=4\left(\frac{\sqrt{3}}{2}.\cos x-\sin x\times\frac{1}{2}\right).\sin x.\ \left(\frac{\sqrt{3}}{2}.\cos x+\sin x\times\frac{1}{2}\right)$$

$$= 4 sin x.(3/4. cos^2x - sin ^2x\times1/4)$$

$$= sin x.(3 cos^2x - sin ^2x)$$

$$= sin x [cos^2 x - sin^2 x + 2.cos^2 x]$$

$$= sin x.[cos^2 x - sin^2 x] + cos x.2 sin x.cos x$$

$$= sin x. cos 2x + cos x.sin 2x$$

$$= sin (x + 2x)$$

$$= sin 3x$$ 

Similarly, $$4\cos x.\cos\left(60+x\right).\cos\left(60-x\right)=\cos3x\ .$$

So, $$\frac{(32\cos^6x-48\cos^4x+18\cos^2x-1)}{[4\sin x\cos x\sin(60-x)\cos(60-x)\sin(60+x)\cos(60+x)]}$$

$$=\frac{4\times2\times\cos6x}{2\times\sin3x\times\cos3x}$$

$$=8\cot6x\ .$$

C is correct choice.

Question 78 Show answer

What is the value of $$\frac{\cos 2A + 2\cos^2 A - 2\cos 2A \cos A}{\sin 2A - 2\sin^2 A \sin 2A }$$?

A

$$2 \cot A$$

B

$$2 \tan A$$

C

$$\cot A$$

D

$$\tan A$$

Question 79 Show answer

What is the value of $$\left[ \frac{(\tan 5\theta + \tan 3\theta)}{4} \cos 4\theta (\tan 5\theta - \tan 3\theta)\right] $$?

A

$$\sin 2\theta$$

B

$$\cos 2\theta$$

C

$$\tan 4\theta$$

D

$$\cot 2\theta$$

Question 80 Show answer

What is the value of $$\left\{\frac{[\sin (x + y) - 2 \sin x + \sin (x - y)]}{[\cos (x - y) + \cos (x + y) - 2 \cos x]}\right\} \times \left[\frac{(\sin 10x - \sin 8x)}{(\cos 10x + \cos 8x)}\right]$$?

A

0

B

$$\tan^2 x$$

C

1

D

$$2 \tan x$$

Question 81 Show answer

If $$\sec\theta+\tan\theta=p,\left(p>1\right)$$ then $$\frac{\operatorname{cosec}\theta+1}{\operatorname{cosec}\theta-1} = ?$$

A

$$\frac{p+1}{p-1}$$

B

$$p^2$$

C

$$\frac{p-1}{p+1}$$

D

$$2p^2$$

View Explanation Hide Explanation

Solution

$$\sec\theta+\tan\theta=p$$ ----(1)
$$\sec\theta-\tan\theta = \frac{1}{p}$$ ----(2)
From eq (1) and (2),
$$2\sec\theta = p + \frac{1}{p}$$
$$\sec\theta = \frac{p^2 + 1}{2p}$$ = $$\frac{hypotenuse}{base}$$
By the Pythagoras theorem,
$${hypotenuse}^2 = {base}^2 + {perpendicular}^2$$
$${p^2 + 1}^2 = {2p}^2 + {perpendicular}^2$$
$$perpendicular = \sqrt{p^4 + 1 - 2p^2}$$
$$perpendicular = {p^2 - 1}^2$$
Now,
$$\frac{\operatorname{cosec}\theta+1}{\operatorname{cosec}\theta-1}$$
$$\frac{\frac{p^2 + 1}{p^2 - 1}+1}{\frac{p^2 + 1}{p^2 - 1}-1}$$
$$\frac{p^2 + 1 + p^2 - 1}{p^2 + 1 - p^2 + 1}$$ = $$p^2$$

Question 82 Show answer

$$\sqrt{\frac{\cot\theta+\cos\theta}{\cot\theta-\cos\theta}}$$ is equal to:

A

$$\sec\theta+\tan\theta$$

B

$$1 + \sec\theta\tan\theta$$

C

$$1 - \sec\theta\tan\theta$$

D

$$\sec\theta-\tan\theta$$

View Explanation Hide Explanation

Solution

$$\sqrt{\frac{\cot\theta+\cos\theta}{\cot\theta-\cos\theta}}$$
= $$\sqrt{\frac{cos\theta(\frac{1}{sin\theta}+1)}{cos\theta(\frac{1}{sin\theta} - 1)}}$$
= $$\sqrt{\frac{1 + sin\theta}{1 - sin\theta} \times \frac{1 + sin\theta}{1 + sin\theta}}$$
= $$\sqrt{\frac{(1 + sin\theta)^2}{1 - sin^2\theta}}$$
= $$\frac{1 + sin\theta}{cos\theta}$$
= $$sec\theta + tan\theta$$

Question 83 Show answer

What is the value of $$\cos 15$$° + $$\cos 105$$°?

A

$$\surd3$$

B

$$\frac{1}{\surd2}$$

C

$$\frac{\surd3}{2}$$

D

$$\frac{1}{\surd3}$$

Question 84 Show answer

What is the value of cos 15° - cos 165°?

A

$$\frac{\surd3}{\surd2}$$

B

$$\frac{2}{(\surd3 - 1)}$$

C

$$\frac{\surd3 + 1}{\surd2}$$

D

$$\frac{\surd3 + 1}{2}$$

View Explanation Hide Explanation

Solution

cos 15 = cos (45-30)

=> cos 15 = cos 45 * cos 30 + sin 45 * sin 30

=> cos 15 = (1/√2)* (√3/2) + (1/√2)*(1/2)

=> cos 15 = (√3 + 1 )/2√2

$$\cos15^{\circ\ }-\cos165^{\circ\ }=\cos15^{\circ\ }+\cos15^{\circ\ }=2\cos15^{\circ\ }.$$

So, 

$$\cos15^{\circ\ }-\cos165^{\circ\ }=\frac{\sqrt{3}+1}{\sqrt{2}}\ .$$

C is correct choice.

Question 85 Show answer

What is the value of $$\frac{\left[2 \cot \times \frac{(p - A)}{2}\right]}{\left[1 + \tan^2 \times \frac{(2p - A)}{2}\right]}$$?

A

$$2 \sin^2 \frac{A}{2}$$

B

$$\cos A$$

C

$$\sin A$$

D

$$2 \cos^2 \frac{A}{2}$$

Question 86 Show answer

What is the value of $$\left(\frac{4}{3}\right) \cot^2 \left(\frac{p}{6}\right) + 3 \cos^2 (150^\circ)$$ $$- 4 \cosec^2 45^\circ$$ + $$8 \sin \left(\frac{p}{2}\right)$$?

A

$$\frac{25}{4}$$

B

1

C

$$-\frac{7}{2}$$

D

$$\frac{13}{2}$$

View Explanation Hide Explanation

Solution

p=pi
so we get after sustituting :
12+9/4-16+8
4+9/4 
=25/4

Question 87 Show answer

What is the value of $$\sin$$ 75° + $$\sin$$ 15°?

A

$$\surd3$$

B

$$2\surd3$$

C

$$\surd(\frac{3}{2})$$

D

$$\frac{3}{\surd2}$$

Question 88 Show answer

What is the value of 

$$[\sin (90$$° - $$10\theta) - \cos (p - 6\theta)]/[\cos (\frac{p}{2} - 10\theta) - \sin (p - 6\theta)]$$?

A

$$\tan 2\theta$$

B

$$\cot 2\theta$$

C

$$\cot \theta$$

D

$$\cot 3\theta$$

Question 89 Show answer

What is the value of $$\tan \left(\frac{\pi}{4} + A\right) \times \tan \left(\frac{3\pi}{4} + A\right)?$$

A

1

B

0

C

$$\cot \frac{A}{2}$$

D

-1

View Explanation Hide Explanation

Solution

We know $$\tan\ \frac{\pi}{4}=1\ ;\ \tan\ \frac{3}{4}\pi\ \ =-1\ $$
and tan (A+B) = (tan A +tan B )/1-tan A tan B
Now using formula we get $$\tan\left(\frac{\pi}{4}+A\ \right)\times\ \tan\left(\frac{3}{4}\pi\ +A\right)$$
=$$\frac{1+\tan\ A}{1-\tan\ A}\times\ \frac{-1+\tan\ A}{1+\tan\ A}$$
=-1

Question 90 Show answer

If $$3(\cot^2 \phi - \cos \phi) = \cos^2 \phi, 0^\circ < \phi < 90^\circ$$, then the value of $$(\tan^2 \phi + \cosec^2 \phi + \sin^2 \phi)$$ is:

A

$$\frac{13}{3}$$

B

$$\frac{61}{12}$$

C

$$\frac{25}{12}$$

D

$$\frac{15}{4}$$

View Explanation Hide Explanation

Solution

$$3(\cot^2 \phi - \cos \phi) = \cos^2 \phi$$
Put the value of $$\phi = 60\degree$$,
$$3(\cot^2 60\degree - \cos 60\degree) = \cos^2 60\degree$$
$$3(\frac{1}{3} - \frac{1}{4}) = \frac{1}{4}$$
$$3(\frac{1}{12}) = \frac{1}{4}$$
$$\frac{1}{4} = \frac{1}{4}$$
Now,
$$(\tan^2 \phi + \cosec^2 \phi + \sin^2 \phi)$$
Put the value of $$\phi = 60\degree$$,
= $$(\tan^260\degree + \cosec^260\degree + \sin^2 60\degree)$$
= $$3 + \frac{4}{3} + \frac{3}{4}$$
= $$\frac{36 + 16 + 9}{12} = \frac{61}{12}$$

Question 91 Show answer

If $$5\sin\theta-4\cos\theta=0,0^{\circ}<\theta<90^{\circ}$$, then the value of $$\frac{5\sin\theta-2\cos\theta}{5\sin\theta+3\cos\theta}$$is:

A

$$\frac{3}{8}$$

B

$$\frac{3}{7}$$

C

$$\frac{2}{7}$$

D

$$\frac{5}{8}$$

View Explanation Hide Explanation

Solution

$$5\sin\theta-4\cos\theta=0,0^{\circ}<\theta<90^{\circ}$$
= $$5\sin\theta = 4\cos\theta$$
= $$\tan\theta = \frac{4}{5}$$
Now,
$$\frac{5\sin\theta-2\cos\theta}{5\sin\theta+3\cos\theta}$$
= $$\frac{\cos\theta(5\tan\theta-2)}{\cos\theta(5\tan\theta+3)}$$
= $$\frac{5\tan\theta-2}{5\tan\theta+3}$$
= $$\frac{5\times \frac{4}{5} -2}{5\times \frac{4}{5} +3}$$
= $$\frac{4 -2}{4 +3}$$
= $$\frac{2}{7}$$

Question 92 Show answer

If $$P + Q + R = 60$$°, then what is the value of $$\cos Q \cos R (\cos P - \sin P) + \sin Q \sin R (\sin P - \cos P)$$?

A

$$\frac{1}{2}$$

B

$$\frac{\surd3}{2}$$

C

$$\frac{1}{\surd2}$$

D

$$\surd2$$

Question 93 Show answer

If $$\sec \theta (\cos \theta + \sin \theta) = \surd2$$, then what is the value of $$\frac{(2 \sin \theta)}{(\cos \theta - \sin \theta)}$$?

A

$$3\surd2$$

B

$$\frac{3}{\surd2}$$

C

$$\frac{1}{\surd2}$$

D

$$\surd2$$

View Explanation Hide Explanation

Solution

$$\sec\theta(\cos\theta+\sin\theta)=\surd2$$

or, $$(1+\tan\theta)=\surd2\ .$$

or,$$\theta=\tan^{-1}\left(\sqrt{2}-1\right)=22.5^{\circ\ }\ .$$

So,

$$\frac{(2\sin\theta)}{(\cos\theta-\sin\theta)}\ $$

$$=\frac{(2\sin22.5^{\circ\ })}{(\cos22.5^{\circ\ }-\sin22.5^{\circ\ })}\ $$

$$=1.41=\sqrt{2}\ .$$

D is correct choice.

Question 94 Show answer

If $$\sin (A - B) = \frac{1}{2}$$ and $$\cos (A + B) = \frac{1}{2}$$, then what is the value of $$\sin A \cos A + \sin^2 A \sin B \cos B + \cos^3 A \cos B \tan A$$?

A

$$\frac{1}{2}$$

B

$$\frac{3}{4}$$

C

$$\frac{1}{4}$$

D

$$\frac{3}{2}$$

Question 95 Show answer

If $$\tan \theta + \sec \theta = \frac{(x - 2)}{(x + 2)}$$, then what is the value of $$\cos \theta$$?

A

$$\frac{(x^2 - 1)}{(x^2 + 1)}$$

B

$$\frac{(2x^2 - 4)}{(2x^2 + 4)}$$

C

$$\frac{(x^2 - 4)}{(x^2 + 4)}$$

D

$$\frac{(x^2 - 2)}{(x^2 + 2)}$$

View Explanation Hide Explanation

Solution

$$\tan\theta+\sec\theta=\frac{(x-2)}{(x+2)}$$

we know, $$\left(\tan\theta+\sec\theta\right)\left(\sec\theta\ -\tan\theta\ \right)=1\ .$$

So, $$\left(\sec\theta\ -\tan\theta\ \right)=\frac{x+2}{x-2}\ .$$

So, $$2\sec\theta\ =\frac{x+2}{x-2}\ +\frac{x-2}{x+2}=\frac{x^2+4x+4+x^2-4x+4}{x^2-4}=\frac{2\left(x^2+4\right)}{x^2-4}\ .$$

or, $$\cos\theta\ =\frac{x^2-4}{x^2+4}\ .$$

So, C is correct choice.

Question 96 Show answer

The value of $$\sin^2 48^\circ + \sin^2 42^\circ - \sec^2 30^\circ + \tan^2 60^\circ $$ is equal to:

A

2

B

$$\frac{7}{3}$$

C

$$\frac{8}{3}$$

D

$$\frac{5}{3}$$

View Explanation Hide Explanation

Solution

$$\sin 42^\circ$$ = $$\sin (90 - 42)^\circ$$ = $$\cos 48^\circ$$

$$\sin^2 48^\circ + \cos^2 48^\circ - \sec^2 30^\circ + \tan^2 60^\circ$$ = 1 - $$\frac{4}{3}$$ + 3 =$$\frac{8}{3}$$

(Since we know that $$sin ^2 \theta + cos ^2\theta = 1, tan 60^o = \sqrt{3}, sec 60^o = \frac{2}{\sqrt{3}}$$)

Question 97 Show answer

What is the value of $$[(\cos 3\theta + 2\cos 5\theta + \cos 7\theta) \div (\cos \theta + 2\cos 3\theta + \cos 5\theta)] + \sin 2\theta \tan 3\theta$$?

A

$$\cos 2\theta$$

B

$$\sin 2\theta$$

C

$$\tan 2\theta$$

D

$$\cot \theta \sin 2\theta$$

Question 98 Show answer

What is the value of $$\left[(\sec 2\theta + 1)\sqrt{\sec^2 \theta -1}\right] \times \frac{1}{2} (\cot \theta - \tan \theta)$$

A

0

B

1

C

$$\cosec \theta$$

D

$$\sec \theta$$

View Explanation Hide Explanation

Solution

we know $$\sqrt{\ \sec^2\theta\ -1}=\tan\theta\ $$
Now $$\frac{1}{2}\left(\cot\ \theta\ -\tan\theta\ \right)=\frac{\frac{1}{2}\left(\cos^2\theta\ -\sin^2\theta\ \right)}{\sin\theta\ \cos\theta\ }=\frac{\cos2\theta}{\sin2\theta\ }$$
So we get 
$$\left(\sec2\theta\ +1\right)\times\ \tan\theta\ \times\ \frac{\cos2\theta\ }{\sin2\theta\ }$$
Multiplying we get :
$$\frac{\tan\theta\ }{\sin2\theta\ }+\frac{\tan\theta\ \cos2\theta\ }{\sin2\theta\ }$$
we get $$\frac{\tan\theta\ }{\sin2\theta\ }\left(1+\cos2\theta\ \right)\ $$
we get $$\frac{\tan\theta\ }{2\sin\theta\ \cos\theta\ }\times\ 2\cos^2\theta\ \ $$
=1

Question 99 Show answer

What is the value of $$\sin (B - C) \cos (A - D) + \sin (A - B) \cos (C - D) + \sin (C - A) \cos (B - D)?$$

A

$$\frac{3}{2}$$

B

-3

C

1

D

0

Question 100 Show answer

If $$ (1 + \tan^2 \theta) + (1 + ( \tan^2 \theta)^{-1}) = k,  then  \sqrt k = ?$$

A

$$\cosec \theta \cos \theta$$

B

$$\sin \theta \cos \theta$$

C

$$\cosec \theta \sec \theta$$

D

$$\sin \theta \sec \theta$$

View Explanation Hide Explanation

Solution

$$ (1 + \tan^2 \theta) + (1 + ( \tan^2 \theta)^{-1})$$ = k

= where θ is replaced by  x

use this identity to solve this problem tanx = $$\frac{sinx}{cosx}$$

from the question

= (1+$$\frac{sin^2x}{cos^2x}$$) + (1+$$\frac{1}{tan^2x}$$) = k

=  ($$\frac{cos^2x+sin^2x}{cos^2x}$$)+(1+$$\frac{1}{sin^2x÷cos^2x}$$) = k

= use this trigonometric identity $$\cos^2 \theta + \sin^2 \theta$$ = 1

= ($$\frac{1}{cos^2x}$$) + (1+$$\frac{cos^2x}{sin^2x}$$) = k

= ($$\frac{1}{cos^2x}$$) + ($$\frac{sin^2x+cos^2x}{sin^2x}$$) = k

= ($$\frac{1}{cos^2x}$$) + ($$\frac{1}{sin^2x}$$) take the order of this then

= on further simplification it becomes

= ($$\frac{cos^2x+sin^2x}{cos^2x × sin^2x}$$) = k

= ($$\frac{1}{cos^2x × sin^2x}$$) = k

we need to find √k hence after squaring we get = $$\frac{1}{cosx × sinx}$$

where  $$\frac{1}{cosx}$$ = secx  AND   $$\frac{1}{sinx}$$   =  cosecx from the trigonometric inverse functions

 Hence the solution is  cosecx secx

Question 101 Show answer

If $$\cosec \theta = \frac{13}{12},  then  \sin \theta + \cos \theta - \tan \theta$$ is equal to:

A

$$\frac{139}{65}$$

B

$$\frac{91}{65}$$

C

$$-\frac{71}{65}$$

D

$$\frac{71}{65}$$

View Explanation Hide Explanation

Solution

$$\cosec \theta =\frac{H}{P}= \frac{13}{12}$$

H= 13, P= 12, B= 5 (We know that $$hypotnuse^2 = side^2 + side^2.  So we will get  B = 5$$

$$\sin \theta + \cos \theta - \tan \theta$$

$$\frac{P}{H}+\frac{B}{H}-\frac{P}{B}$$

$$\frac{12}{13}+\frac{5}{13}-\frac{12}{5}=-\frac{71}{65}$$

Question 102 Show answer

If $$\cot \theta = \sqrt7$$, then the value of $$\frac{\cosec^2 \theta - \sec^2 \theta}{\cosec^2 \theta + \sec^2 \theta}$$ is:

A

$$\frac{7}{9}$$

B

$$\frac{3}{4}$$

C

$$\frac{8}{9}$$

D

$$\frac{2}{3}$$

Question 103 Show answer

If $$\frac{\cos^2 \theta}{\cot^2 \theta - \cos^2 \theta} = 3, 0^\circ < \theta < 90^\circ$$, then the value of $$\cot \theta + \cosec \theta$$ is:

A

$$\sqrt3$$

B

$$\frac{\sqrt3}{2}$$

C

$$2\sqrt3$$

D

$$3\frac{\sqrt3}{4}$$

View Explanation Hide Explanation

Solution

Given that,

$$\frac{\cos^2 \theta}{\cot^2 \theta - \cos^2 \theta} = 3$$

$$\Rightarrow \cos^2 \theta=3( \cot^2\theta-\cos^2 \theta)$$

$$\Rightarrow \cos^2 \theta +3 \cos^2 \theta=3\cot^2\theta$$

$$\Rightarrow 4\cos^2 \theta =3\cot^2\theta$$

$$\Rightarrow 4\cos^2 \theta =3\dfrac{\cos^2 \theta}{\sin^2 \theta}$$

$$\Rightarrow \sin^2 \theta=\dfrac{3}{4}$$

$$\Rightarrow \sin^2 \theta=\dfrac{3}{4}$$

$$\Rightarrow \sin \theta=\dfrac{\sqrt 3}{2}=\sin 60$$

$$\theta=60$$

Now, substituting the values,

$$\Rightarrow \cot \theta + \cosec \theta =\cot 60^\circ+\cosec 60^\circ$$

$$\Rightarrow \cot 60^\circ+\cosec 60^\circ=\dfrac{1}{\sqrt 3}+\dfrac{2}{\sqrt 3}=\dfrac{3}{\sqrt3}=\sqrt{3}$$

Question 104 Show answer

What is the value of (cos 40° - cos 140°)/(sin 80° + sin 20°)?

A

$$2\surd3$$

B

$$\frac{2}{\surd3}$$

C

$$\frac{1}{\surd3}$$

D

$$\surd3$$

View Explanation Hide Explanation

Solution

Solve the numerator,

⇒ cos 40° - cos140° = - 2 sin[(140° + 40°)/2] × sin[(40° - 140°)/2]

⇒ cos 40° - cos140° = 2 sin[(140° + 40°)/2] × sin[(140° - 40°)/2]

⇒ cos 40° - cos 140° = 2sin90°.sin50°

⇒ cos 40° - cos 140° = 2sin50°

Solve the denominator,

sin 80° + sin 20° = 2 sin[(80° + 20°)/2] × cos[(80° - 20°)/2]

⇒ sin 80° + sin 20° = 2sin50°.cos30°

⇒ sin80° + sin20° = 2sin50° × (√3/2)

Replacing the respective value in the given equation:

(cos40° - cos140°)/(sin80° + sin20°) = (2sin50°)/(2sin50° × √3/2) = 2/√3

B is correct choice.

Question 105 Show answer

What is the value of $$\cos(90 + 75$$°)?

A

$$2+\surd3$$

B

$$-2-\surd3$$

C

$$\surd3+1$$

D

$$-\surd3-1$$

Question 106 Show answer

What is the value of $$\frac{[1 - \tan(90 - \theta)]^2}{[\cos^2(90 - \theta)]} - 1$$?

A

$$-\sin 2\theta$$

B

$$-\cos 2\theta$$

C

$$\cos 2\theta$$

D

$$\sin 2\theta$$

View Explanation Hide Explanation

Solution

$$\frac{[1-\tan(90-\theta)]^2}{[\sec^2(90-\theta)]}-1$$

$$=\frac{[1-\cot(\theta)]^2}{[\cosec^2(\theta)]}-1\ .$$

$$=[1-\frac{2\cos(\theta)}{\sin\left(\theta\ \right)}.\ \frac{\sin^2\left(\theta\ \right)}{1}-1\ .$$

$$=-\sin\left(2\theta\ \right)\ .$$

A is correct choice.

Question 107 Show answer

What is the value of $$\frac{1}{\sin^4 (90 - \theta)}+\frac{1}{[\cos^2 (90 - \theta)] - 1}?$$

A

$$\tan^2 \theta sec^2 \theta$$

B

$$\sec^4 \theta$$

C

$$\tan^4 \theta$$

D

$$\tan^2 \theta \sin^2 \theta$$

View Explanation Hide Explanation

Solution

$$\frac{1}{\sin^4(90-\theta)}+\frac{1}{[\cos^2(90-\theta)]-1}\ .$$

$$=\frac{1}{\cos^4\theta}-\frac{1}{1-\sin^2\theta\ }\ .$$

$$=\frac{1}{\cos^4\theta}-\frac{1}{\cos^2\theta\ }\ .$$

$$=\frac{1-\cos^2\theta\ }{\cos^4\theta}\ .$$

$$=\frac{\sin^2\theta\ }{\cos^4\theta}\ .$$

$$=\tan^2\theta\ \sec^2\theta\ .$$

A is correct choice.

Question 108 Show answer

What is the value of $$\frac{[2 \sin (45 + \theta) \sin (45 - \theta)]}{\cos 2\theta}$$?

A

0

B

$$\tan 2\theta$$

C

$$\cot 2\theta$$

D

1

Question 109 Show answer

What is the value of $$\frac{\left\{[4 \cos (90 - A) \sin^3(90 + A)] - [4 \sin (90 + A) \cos^3(90 - A)]\right\}}{\cos\left[\frac{180 + 8A}{2}\right]}$$?

A

1

B

-1

C

0

D

2

Question 110 Show answer

What is the value of sin $$(630$$° + $$A) + \cos A$$?

A

$$\frac{\surd3}{2}$$

B

$$\frac{1}{2}$$

C

0

D

$$\frac{2}{\surd3}$$

Question 111 Show answer

If $$(A + B + C) = 90$$°, then what is the value of $$\sin \left(\frac{A}{2}\right) \sin \left[\frac{(180 — B — C)}{2}\right] + \cos \left(\frac{A}{2}\right) \sin \frac{(B + C)}{2}$$?

A

$$\frac{1}{2}$$

B

1

C

$$\frac{1}{\surd2}$$

D

$$\frac{\surd3}{2}$$

View Explanation Hide Explanation

Solution

We have :
A+B+C =90 
Now (180-B-C)/2 = 90-(B-C)/2
Now sin (180-B-C)/2)= sin (90-(B+C)/2 )= cos (B+C)/2
So we get the expression as
$$\sin\ \frac{A}{2}\cos\ \frac{B+C}{2}+\cos\ \frac{A}{2}\sin\ \frac{B+C}{2}=\sin\ \left(\frac{A+B+C}{2}\right)\ =\sin\ 45\ =\ \frac{1}{\sqrt{\ 2}}$$

Question 112 Show answer

If $$\sec \theta - \tan \theta = P$$, then $$\cosec \theta = ?$$

A

$$\frac{2P}{1 - P^2}$$

B

$$\frac{1 - P^2}{1 + P^2}$$

C

$$\frac{P^2 + 1}{1 - P^2}$$

D

$$\frac{2P}{1 + P^2}$$

View Explanation Hide Explanation

Solution

As per the given question,

$$\Rightarrow \sec \theta - \tan \theta = P$$

$$\Rightarrow \dfrac{1}{\cos \theta} - \dfrac{\sin \theta}{\cos\theta} = P$$

$$\Rightarrow \dfrac{1-\sin \theta}{\cos \theta} = P$$

Squaring both side of the given equation,

$$\Rightarrow \dfrac{(1-\sin \theta)^2}{(\cos \theta)^2} = P^2$$

$$\Rightarrow \dfrac{(1-\sin \theta)^2}{1-\sin^2\theta} = P^2$$

$$\Rightarrow \dfrac{(1-\sin \theta)^2}{(1-\sin\theta)(1+\sin\theta)} = P^2$$

$$\Rightarrow \dfrac{(1-\sin \theta)}{(1+\sin\theta)} = P^2$$

$$\Rightarrow 1-\sin \theta = P^2(1+\sin\theta)$$

$$\Rightarrow 1-\sin \theta = P^2+P^2\sin\theta$$

$$\Rightarrow \sin\theta(P^2+1)=1-P^2$$

$$\Rightarrow \sin\theta=\dfrac{1-P^2}{(P^2+1)}$$

Hence $$\cosec \theta=\dfrac{P^2+1}{1-P^2}$$

Question 113 Show answer

If $$\tan^2 \theta - 3 \sec \theta + 3 = 0, 0^\circ < \theta < 90^\circ$$, then the value of $$ \sin \theta  + \cot \theta$$ is:

A

$$\frac{5\sqrt3}{3}$$

B

$$\frac{5\sqrt3}{6}$$

C

$$2\sqrt3$$

D

$$3\sqrt3$$

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Solution

We know
$$\tan^2\theta\ =\sec^2\theta\ -1$$
substituting we get 
$$\sec^2\theta\ -3\sec\theta\ +2\ =0$$
we get $$\sec\theta\ =2$$and $$\theta\ =60$$
so $$\sin\theta\ +\cot\theta\ =\frac{5\sqrt{\ 3}}{6}$$

Question 114 Show answer

The value of $$\sin^2 64^\circ + \cos 64^\circ \sin 26^\circ + 2 \cos 43^\circ \cosec 47^\circ$$ is:

A

4

B

1

C

2

D

3

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Solution

$$\sin^2 64^\circ + \cos 64^\circ \sin 26^\circ + 2 \cos 43^\circ \cosec 47^\circ$$
= $$\sin^2 64^\circ + \cos 64^\circ \sin(90 - 64^\circ) + 2 \cos 43^\circ \cosec(90 - 43^\circ)$$
= $$\sin^2 64^\circ + \cos^2 64^\circ  + 2 \cos 43^\circ se43^\circ$$
= $$1 + 2$$ = 3

Question 115 Show answer

What is the value of 

$$\cos [\frac{(180 - \theta)}{2}] \cos [\frac{(180 - 9\theta)}{2}] + \sin [\frac{(180 - 3\theta)}{2}] \sin [\frac{(180 - 13\theta)}{2}]?$$

A

$$sin 2\theta \sin 4\theta$$

B

$$\cos 2\theta \cos 6\theta$$

C

$$\sin 2\theta \sin 6\theta$$

D

$$\cos 2\theta \cos 4\theta$$

Question 116 Show answer

What is the value of $$\frac{[1 +2 \cot^2(90 - x) - 2\cosec(90 - x) \cot(90 - x)]}{[\cosec(90 - x) - \cot(90 - x)]}$$?

A

$$\cos x + \sin x$$

B

$$\sin x - \cos x$$

C

$$\sec x + \tan x$$

D

$$\sec x - \tan x$$

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Solution

$$\frac{[1+2\cot^2(90-x)-2\operatorname{cosec}(90-x)\cot(90-x)]}{[\operatorname{cosec}(90-x)-\cot(90-x)]}$$

$$=\frac{[1+2\tan^2(x)-2\sec(x)\tan(x)]}{[\sec(x)-\tan(x)]}$$

$$=\frac{[\sec^2x-\tan^2x+2\tan^2(x)-2\sec(x)\tan(x)]}{[\sec(x)-\tan(x)]}$$

$$=\frac{[\sec\left(x\right)-\tan\left(x\right)]^2}{[\sec(x)-\tan(x)]}\ .$$

$$=[\sec\left(x\right)-\tan\left(x\right)]\ .$$

D is correct choice.

Question 117 Show answer

What is the value of $$\frac{[1 - \tan (90 - \theta) + \sec (90 - \theta)]}{[\tan (90 - \theta) + \sec (90 - \theta) + 1]}$$?

A

$$\cot (\frac{θ}{2})$$

B

$$\tan (\frac{θ}{2})$$

C

$$\sin \theta$$

D

$$\cos \theta$$

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Solution

$$\frac{[1-\tan(90-\theta)+\sec(90-\theta)]}{[\tan(90-\theta)+\sec(90-\theta)+1]}$$

$$=\frac{[1-\cot\theta\ +\operatorname{cosec}\theta]}{[\cot\theta+\operatorname{cosec}\theta+1]}\ .$$

$$=\frac{\sin\theta\ -\cos\theta\ +1}{\sin\theta\ +1+\cos\theta\ }\ .$$

Now, 

$$\sin\theta\ +1-\cos\theta\ =\frac{2\tan\frac{\theta}{2}+1+\tan^2\frac{\theta}{2}\ -1+\tan^2\frac{\theta\ }{2}\ \ }{1+\tan^2\ \frac{\theta}{2}\ }$$

And, 

$$\sin\theta\ +1+\cos\theta\ =\frac{2\tan\frac{\theta}{2}+1+\tan^2\frac{\theta}{2}\ +1-\tan^2\frac{\theta\ }{2}\ \ }{1+\tan^2\ \frac{\theta}{2}\ }$$

So, 

$$\frac{[1-\tan(90-\theta)+\sec(90-\theta)]}{[\tan(90-\theta)+\sec(90-\theta)+1]}$$

$$=\frac{2\tan\frac{\theta}{2}\left(1+\tan\frac{\theta}{2}\right)}{2\left(\tan\frac{\theta}{2}+1\right)}\ .$$

$$=\tan\frac{\theta}{2}\ .$$

B is correct choice.

Question 118 Show answer

What is the value of $$\frac{[\tan (90 - A) + \cot (90 - A)]^2}{[2 \sec^2 (90 - 2A)]}$$?

A

0

B

1

C

2

D

-1

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Solution

$$\frac{[\tan(90-A)+\cot(90-A)]^2}{[2\sec^2(90-2A)]}$$

$$=\frac{[\cot(A)+\tan(A)]^2}{\frac{2}{\sin^2\left(2A\right)}}\ .$$

$$=\frac{\left[\frac{\cos(A)}{\sin\left(A\right)}+\frac{\sin(A)}{\cos\left(A\right)}\right]^2}{\frac{2}{4\left(\sin A\ \cos A\right)^2}}\ .$$

$$=2\left[\frac{\cos(A)}{\sin\left(A\right)}+\frac{\sin(A)}{\cos\left(A\right)}\right]^2\left(\sin\left(A\right)\cos\left(A\right)\right)^2\ .$$

$$=2\left[\cos^2(A)+\sin^2(A)\right]^2\ .$$

$$=2\left(1\right)^2\ .$$

$$=2\ .$$

C is correct choice.

Question 119 Show answer

What is the value of $$[(\sin 59$$° $$\cos 31$$° + $$\cos 59$$° $$\sin 31$$°) $$\div$$ ($$\cos 20$$° $$\cos 25$$° - $$\sin 20$$° $$\sin 25$$°$$)]$$?

A

$$\frac{1}{\surd2}$$

B

$$\frac{2}{\surd2}$$

C

$$\surd3$$

D

$$\surd2$$

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Solution

sinAcosb + sinBcosA= sin(A+B)

cosAcosB - cosAcosB= cos(A+B)

So, $$sin(A+B)\div cos(A-B)=sin(59+31)\div cos(25+20)$$= $$sin 90\div cos 45$$= $$1\div1\div\sqrt{2}=\sqrt{2}$$

Therefore, Option D is correct.

Question 120 Show answer

What is the value of sin (90° + 2A)[4 - $$\cos^2$$ (90° - 2A)]?

A

$$2(\cos^3 A - \sin^3 A)$$

B

$$2(\cos^3 A + \sin^3 A)$$

C

$$4(\cos^6 A+ \sin^6 A)$$

D

$$4(\cos^6 A - \sin^6 A)$$

Question 121 Show answer

If $$\cos \theta = \frac{2p}{1 + p^2}$$, then $$\tan \theta$$ is equal to:

A

$$\frac{1 - p^2}{2p}$$

B

$$\frac{p^2}{1 + p^2}$$

C

$$\frac{1 - p^2}{1 + p^2}$$

D

$$\frac{2p}{1 - p^2}$$

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Solution

Given that,

$$\cos \theta = \frac{2p}{1 + p^2}$$

$$\sin \theta=\sqrt{1-\cos^2\theta}=\sqrt{1-(\frac{2p}{1 + p^2})^2}=\sqrt{\dfrac{(1+p^2)^2-4p^2}{(1+p^2)^2}}$$

$$\sin \theta=\sqrt{\dfrac{1+p^4+2p^2-4p^2}{(1+p^2)^2}}=\sqrt{\dfrac{1+p^4-2p^2}{(1+p^2)^2}}=\sqrt{\dfrac{(1-p^2)^2}{(1+p^2)^2}}=\dfrac{1-p^2}{1+p^2}$$

So, $$\tan \theta=\dfrac{\dfrac{1-p^2}{1+p^2}}{ \frac{2p}{1 + p^2}}=\dfrac{1-p^2}{2p}$$

Question 122 Show answer

If $$\tan 4 \theta = \cot (2\theta + 30^\circ)$$, then $$\theta$$ is equal to:

A

$$20^\circ$$

B

$$25^\circ$$

C

$$15^\circ$$

D

$$10^\circ$$

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Solution

Given that,

$$\tan 4 \theta = \cot (2\theta + 30^\circ)$$

We know that $$\cot (90-A)=\tan A$$

Hence,

$$\Rightarrow \cot(90^\circ-4\theta)=\cot (2\theta +30^\circ)$$

$$\Rightarrow 90^\circ -4\theta=2\theta + 30^\circ$$

$$\Rightarrow 6\theta=90^\circ -30^\circ$$

$$\Rightarrow \theta=\dfrac{60^\circ}{6}=10^\circ$$

Question 123 Show answer

What is the value of $$[\cos (90 + A)\div \sec (270 - A)] + [\sin (270 + A)\div \cosec (630 - A)]$$?

A

$$3 \sec A$$

B

$$\tan A \sec A$$

C

0

D

1

Question 124 Show answer

What is the value of $$\cos (90 - B) \sin (C - A) + \sin (90 + A) \cos (B + C) - \sin (90 - C) \cos (A + B)$$?

A

1

B

$$\sin ( A + B - C)$$

C

$$\cos ( B + C - A)$$

D

0

Question 125 Show answer

What is the value of $$\cot (90 - x) \sin^4 (90 - x) + \cot (180 - x) \sin^4 (180 - x)$$?

A

$$\frac{(\cos 4x)}{4}$$

B

$$\frac{(\sin^2 2x)}{2}$$

C

$$\frac{(\cos^2 2x)}{2}$$

D

$$\frac{(\sin 4x)}{4}$$

Question 126 Show answer

What is the value of $$\frac{[\sin (90 - A) + \cos (180 - 2A)]}{[\cos (90 - 2A) + \sin (180 - A)]}$$?

A

$$\sin (\frac{A}{2}) \cos A$$

B

$$\cot (\frac{A}{2})$$

C

$$\tan (\frac{A}{2})$$

D

$$\sin A \cos (\frac{A}{2})$$

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Solution

$$\frac{[\sin(90-A)+\cos(180-2A)]}{[\cos(90-2A)+\sin(180-A)]}$$

$$=\frac{[\cos A-\cos2A]}{[\sin(2A)+\sin(A)]}\ .$$

$$=\frac{2.\sin\frac{3A}{2}.\ \sin\frac{A}{2}}{2.\ \sin\frac{3A}{2}.\cos\frac{A}{2}}\ .$$

$$=\tan\frac{A}{2}\ .$$

C is correct choice.

Question 127 Show answer

What is the value of 

$$ \left\{ \sin (90 + x) \cos [\pi - (x - y)] \right\}$$ $$+ \left\{  \cos (90 + x) \sin [\pi - (x - y)] \right\}$$?

A

$$-\cos y$$

B

$$-\sin y$$

C

$$\cos x$$

D

$$\tan y$$

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Solution

We know that : $$Sin\left(A+B\right)=SinACosB+CosASinB\ .$$

So, $$\left\{ \sin (90 +x) \cos [\pi - (x - y)] \right\}$$ $$+ \left\{ \cos (90 + x) \sin [\pi - (x - y)] \right\}$$

$$=Sin\left(\left(90^{\circ\ }+x\right)+\pi\ -\left(x-y\right)\right)\ .$$

$$=Sin\left(90^{\circ\ }+x+180^{\circ\ }-x+y\right)\ .$$

$$=Sin\left(270^{\circ\ }+y\right)\ .$$

$$=-Cos\ y\ .$$

A is correct choice.

Question 128 Show answer

What is the value of $$\sin(180 - \theta) \sin(90 - \theta) - \left[\frac{\cot(90 - \theta)}{1 + \tan^2 \theta}\right]$$

A

$$\cos^2 \theta \sin \theta$$

B

$$\frac{\cot \theta}{1 + \cot^2 \theta}$$

C

$$\frac{\tan \theta}{1 + \tan^2 \theta}$$

D

0.

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Solution

$$\sin(180-\theta)\sin(90-\theta)-\left[\frac{\cot(90-\theta)}{1+\tan^2\theta}\right]$$

$$=\sin\theta\ \cos\theta-\left[\frac{\tan\theta}{\sec^2\theta}\right]$$

$$=\sin\theta\ \cos\theta-\left[\frac{\sin\theta}{\cos\theta}.\ \frac{\cos^2\theta\ }{1}\right]$$

$$=\sin\theta\ \cos\theta-\left[\sin\theta\ \cos\theta\ \right]$$

$$=0\ .$$

D is correct choice.

Question 129 Show answer

What is the value of 

$$[\tan^2 (90 - \theta) - \sin^2 (90 - \theta)] \cosec^2 (90 - \theta) \cot^2 (90 - \theta)?$$

A

0

B

1

C

-1

D

2

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Solution

$$[(\sin^2 (90 - \theta)/\cos^2 (90 - \theta)) - \sin^2 (90 - \theta)] \cos^2 (90 - \theta) /(\sin^4 (90 - \theta))$$

=$$(\sin^2 (90 - \theta)(1-\cos^2 (90 - \theta)/(\cos^2 (90 - \theta))) \cos^2 (90 - \theta) /(\sin^4 (90 - \theta))$$

$$(1-\cos^2 (90 - \theta)$$=$$\sin^2 (90 - \theta)$$

=$$(\sin^4 (90 - \theta)(\cos^2 (90 - \theta)/( \cos^2 (90 - \theta) /(\sin^4 (90 - \theta)))$$

=1

Question 130 Show answer

If $$2 \sin \theta = 5 \cos \theta,  then  \dfrac{\sin \theta + \cos \theta}{\sin \theta - \cos \theta}$$ is equal to:

A

$$\frac{5}{3}$$

B

$$\frac{9}{5}$$

C

$$\frac{2}{3}$$

D

$$\frac{7}{3}$$

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Solution

$$2\sin \theta = 5 \cos \theta$$

=>$$\dfrac{sin \theta}{\cos \theta} = \frac{5}{2}$$

=>$$\dfrac{\sin \theta + \cos \theta}{\sin \theta - \cos \theta}$$ 

Divide both numerator and denominator by $$\cos \theta$$,

= $$\dfrac{\frac{5}{2} + 1}{\frac{5}{2} -1}$$

=$$\dfrac{\frac{7}{2}}{\frac{3}{2}}$$ 

=$$\dfrac{7}{3}$$

So , the answer would be option d) 7:3

Question 131 Show answer

If $$\frac{1}{\operatorname{cosec}\theta-1}+\frac{1}{\operatorname{cosec}\theta+1}=2\sec\theta$$, $$0^\circ < \theta < 90^\circ$$, then the value of $$(\cot \theta + \cos \theta)$$ is:

A

$$\frac{2 + \sqrt3}{\sqrt2}$$

B

$$\frac{2 + \sqrt2}{2}$$

C

$$\frac{1 + \sqrt2}{2}$$

D

$$1 + \sqrt2$$

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Solution

$$\frac{1}{\operatorname{cosec}\theta-1}+\frac{1}{\operatorname{cosec}\theta+1}=2\sec\theta$$

$$\frac{\operatorname{cosec}\theta\ +1+\operatorname{cosec}\theta\ -1}{\operatorname{cosec}^2\theta-1}=2\sec\theta$$

$$\frac{2\operatorname{cosec}\theta\ \ }{\cot^2\theta\ }=2\sec\theta$$

$$\frac{\operatorname{cosec}\theta\ }{\sec\theta\ }\ =\cot^2\theta\ $$

$$\cot\theta\ =\cot^2\theta\ $$

$$\cot^2\theta\ -\cot\theta\ =0$$

$$\cot\theta\ \left(\cot\theta\ -1\right)\ =0$$

$$\Rightarrow$$  $$\cot\theta=0$$  or   $$\cot\theta-1=0$$

$$\Rightarrow$$  $$\cot\theta=0$$  or  $$\cot\theta=1$$

$$\Rightarrow$$  $$\theta=90^{\circ\ }$$  or   $$\theta=45^{\circ\ }$$

Since $$0^\circ < \theta < 90^\circ$$

$$\Rightarrow$$  $$\theta=45^{\circ\ }$$

$$\therefore\ $$ $$\cot\theta+\cos\theta=\cot45^{\circ\ }+\cos45^{\circ\ }$$

$$=1+\frac{1}{\sqrt{2}}$$

$$=\frac{\sqrt{2}+1}{\sqrt{2}}$$

$$=\frac{\left(\sqrt{2}+1\right)\times\sqrt{2}}{\sqrt{2}\times\sqrt{2}}$$

$$=\frac{2 + \sqrt2}{2}$$

Hence, the correct answer is Option B

Question 132 Show answer

On walking 100 metres towards a building in a horizontal line, the angle of elevation of its top changes from 45° to 60°. What will be the height (in metres) of the building?

A

$$50(3 + \surd3)$$

B

$$100(\surd3 + 1)$$

C

$$150$$

D

$$100\surd3$$

Question 133 Show answer

The value of

$$ \sin^2 42^\circ + \sin^2 48^\circ + \tan^2 60^\circ - \cosec 30^\circ$$ is equal to:

A

5

B

3

C

4

D

2

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Solution

$$ \sin^2 42^\circ + \sin^2 48^\circ + \tan^2 60^\circ - \cosec 30^\circ$$

= $$ \sin^2 42^\circ + \cos^2 42^\circ + \tan^2 60^\circ - \cosec 30^\circ$$

= 1 + 3 - 2 = 2

{Using $$\sin^2\theta + \cos^2\theta = 1, \tan60\degree = \sqrt{3} , \cosec30\degree = 2$$} 

So , the answer would be option d)2.

Question 134 Show answer

Two trees are standing along the opposite sides of a road. Distance between the two trees is 400 metres. There is a point on the road between the trees. The angle of depressions of the point from the top of the trees are 45° and 60°. If the height of the tree which makes 45°angle is 200 metres, then what will be the height (in metres) of the other tree?

A

200

B

$$200\surd3$$

C

$$100\surd3$$

D

250

Question 135 Show answer

A tower stands on the top of a building which is 40 metres high. The angle of depression of a point situated on the ground from the top and bottom of the tower are found to be 60° and 45° respectively. What is the height (in metres) of tower?

A

$$20\surd3$$

B

$$30(\surd3 + 1)$$

C

$$40(\surd3 - 1)$$

D

$$50(\surd3 - 1)$$

Question 136 Show answer

Height of a tower is 120 metres. The angle of elevation of the top of tower from a point B is 75°. Point B is on the ground level. What is the distance (in metres) of point B from the base of tower?

A

$$120(2 - \surd3)$$

B

$$180(3 - \surd3)$$

C

$$180(\surd3 - 1)$$

D

$$150(\surd3 - 1)$$

Question 137 Show answer

If $$4 - 2 \sin ^2 \theta - 5 \cos \theta = 0, 0^\circ < \theta < 90^\circ$$, then the value of $$ \sin \theta + \tan \theta$$ is:

A

$$\frac{3\sqrt{2}}{2}$$

B

$$\frac{3\sqrt{3}}{2}$$

C

$$3\sqrt{2}$$

D

$$2\sqrt{3}$$

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Solution

$$4 - 2 \sin ^2 \theta - 5 \cos \theta = 0$$

Using the equation ,

$$ \sin ^2 \theta + \cos ^2 \theta = 1$$ , in above equation ,

$$4 - 2 (1 - \cos ^2 \theta) - 5 \cos \theta = 0$$

$$4 - 2 +  2\cos ^2 \theta - 5 \cos \theta = 0$$

$$2\cos ^2 \theta - 5 \cos \theta +2 = 0$$

$$\cos\theta = \frac{5 \pm \sqrt{25 -16}}{4}$$

$$\cos\theta = \frac{5 \pm 3}{4}$$

$$\cos\theta = 2 or \frac{1}{2}$$

$$\cos\theta  can't  be  more  than 1, so \cos\theta=\frac{1}{2}$$

$$\sin\theta =  \frac{\sqrt{3}}{2}$$

$$\tan\theta = \sqrt{3}$$

$$ \sin \theta + \tan \theta = \frac{\sqrt{3}}{2} + \sqrt{3} = \frac{3\sqrt{3}}{2}$$

So , the answer would be Option b)$$\frac{3\sqrt{3}}{2}$$.

Question 138 Show answer

If $$\frac{\sin^2 \phi - 3 \sin \phi + 2}{\cos^2 \phi} = 1$$ where $$0^\circ < \phi < 90^\circ$$, then what is the value of $$(\cos 2 \phi + \sin 3 \phi + \cosec 2 \phi)$$?

A

$$\frac{2 + \sqrt{3}}{3}$$

B

$$\frac{3 + 4 \sqrt{3}}{6}$$

C

$$\frac{9 + 4 \sqrt{3}}{6}$$

D

$$\frac{3 + 2 \sqrt{3}}{3}$$

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Solution

$$\frac{\sin^2 \phi - 3 \sin \phi + 2}{\cos^2 \phi} = 1$$
On putting the $$\phi = 30\degree$$,
$$\frac{\sin^230\degree - 3 \sin 30\degree + 2}{\cos^2 30\degree} = 1$$
$$\frac{\frac{1}{4} - 3 \times \frac{1}{2} + 2}{ \frac{3}{4}} = 1$$
$$\frac{1 - 6 + 8}{3} = 1 $$
$$\frac{3}{3} = 1 $$
1 = 1
Now,
$$(\cos 2 \phi + \sin 3 \phi + \cosec 2 \phi)$$
On putting the $$\phi = 30\degree$$,
= $$(\cos 2 \times 30\degree + \sin 3 \times  30\degree + \cosec 2 \times 30\degree)$$
= $$(\cos 60\degree + \sin 90\degree + \cosec 60\degree)$$
= $$\frac{1}{2} + 1 + \frac{2}{\sqrt3}$$
= $$\frac{3\sqrt3 + 4}{2\sqrt3}$$
= $$\frac{3\sqrt3 + 4}{2\sqrt3} \times \frac{2\sqrt3}{2\sqrt3}$$
= $$\frac{18 + 8 \sqrt 3}{12}$$
= $$\frac{9 + 4 \sqrt 3}{6}$$

Question 139 Show answer

If $$\sin \theta = \frac{a}{\sqrt{a^2 + b^2}}, 0^\circ < \theta < 90^\circ$$, then the value of $$\sec \theta + \tan \theta$$

A

$$\frac{\sqrt{a^2 + b^2} + a}{b}$$

B

$$\frac{\sqrt{a^2 + b^2} + b}{2a}$$

C

$$\frac{\sqrt{a^2 + b^2} + a}{2b}$$

D

$$\frac{\sqrt{a^2 + b^2} + b}{a}$$

Question 140 Show answer

If $$\tan \theta = \frac{2}{3},  then  \frac{3 \sin \theta - 4 \cos \theta}{3 \sin \theta + 4 \cos \theta}$$ is equal to:

A

$$-\frac{1}{3}$$

B

$$\frac{2}{3}$$

C

$$-\frac{2}{3}$$

D

$$\frac{1}{3}$$

View Explanation Hide Explanation

Solution

$$\tan \theta = \frac{2}{3}$$

$$\frac{prependicular}{base}=\frac{2}{3},hypotenuse=\sqrt{\ 13}$$

$$\sin=\frac{p}{h},\cos=\frac{b}{h}$$

$$ \frac{3 \sin \theta - 4 \cos \theta}{3 \sin \theta + 4 \cos \theta}$$

$$\frac{\left(\frac{6}{\sqrt{\ 3}}-\frac{12}{\sqrt{\ 3}}\right)}{\frac{6}{\sqrt{\ 3}}+\frac{12}{\sqrt{\ 3}}}=-\frac{6}{18}=-\frac{1}{3}$$

Question 141 Show answer

$$\left(\frac{1}{1 + \sin^2 \theta} + \frac{1}{1 + \cosec^2 \theta}\right)$$ = 

A

$$\cosec^2 \theta$$

B

2

C

1

D

$$\sin^2 \theta$$

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Solution

$$\left(\frac{1}{1 + \sin^2 \theta} + \frac{1}{1 + \cosec^2 \theta}\right)$$

where Cosec(t) = $$\frac{1}{sint}$$ transform the expression 

=$$\frac{1}{1+sin^2t}$$ + $$\frac{1}{1+ (1÷sin^2t)}$$

=$$\frac{1}{1+sin^2t}$$ + $$\frac{1}{(sin^2t+1÷sin^2t)}$$

=$$\frac{1}{1+sin^2t}$$ + $$\frac{sin^2t}{sin^2t+1}$$ take the orders of this then reduce simplification is

=$$\frac{1+ sin^2t}{1+sin^2t}$$

= 1

Question 142 Show answer

$$\left(\frac{2\tan30^{\circ}}{1-\tan^230^{\circ}}\right)$$ = 

A

$$\sqrt3$$

B

$$\frac{1}{\sqrt3}$$

C

$$\frac{1}{3}$$

D

3

View Explanation Hide Explanation

Solution

$$\left(\frac{2\tan30^{\circ}}{1-\tan^230^{\circ}}\right)=\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1-\left(\frac{1}{\sqrt{3}}\right)^2}$$

$$=\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1-\frac{1}{3}}$$

$$=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$$

$$=\frac{3}{\sqrt{3}}$$

$$=\sqrt{3}$$

Hence, the correct answer is Option A

Question 143 Show answer

The upper part of a tree broken over by the wind make an angle of 60° with the ground. The distance between the root and the point where top of the tree touches the ground is 25 metres. What was the height (in metres) of the tree?

A

84.14

B

93.3

C

98.25

D

120.24

Question 144 Show answer

If $$\sec 4\theta = \cosec (\theta + 20^\circ)$$, then $$\theta$$ is equal to:

A

$$22^\circ$$

B

$$18^\circ$$

C

$$14^\circ$$

D

$$20^\circ$$

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Solution

$$\sec 4\theta = \cosec (\theta + 20^\circ)$$

$$\operatorname{cosec}\left(90-4\theta\ \right)=\operatorname{cosec}\left(\theta\ +20^{^{\circ\ }}\right)$$

$$90-4\theta\ =\theta\ +20^{^{\circ\ }}=5\theta\ =70^{\circ\ }$$

$$\theta\ =14^{\circ\ }$$

Question 145 Show answer

If $$\sin 3 \theta = \cos (20^\circ - \theta)$$, then $$\theta$$ is equal to:

A

25

B

35

C

28

D

30

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Solution

$$\sin 3 \theta = \cos (20^\circ - \theta)$$

= $$\cos (90 -(\theta + 70))$$

=$$\sin(\theta + 70)$$

$$3\theta = \theta + 70$$

$$\theta = 35\degree$$

So , the answer would be option b)35.                                                 

Question 146 Show answer

Mohit is standing at some distance from a 60 meters tall building. Mohit is 1.8 meters tall. When Mohit walks towards the building, then the angle of elevation from his head becomes 60° from 45°. How much distance (in metres) Mohit covered towards the building?

A

$$18.6(4 - \surd3)$$

B

$$58.2 - 24.6\surd3$$

C

$$19.4(\surd3+1)$$

D

$$19.4(3 - \surd3)$$

Question 147 Show answer

The height of a tower is 300 meters. When its top is seen from top of another tower,then the angle of depression is 60°. The horizontal distance between the bases of the two towers is 120 metres. What is the height (in metres) of the small tower?

A

88.24

B

106.71

C

92.15

D

112.64

Question 148 Show answer

The value of  $$\left[\frac{\sin^2 24^\circ + \sin^2 66^\circ}{\cos^2 24^\circ + \cos^2 66^\circ} + \sin^2 61^\circ + \cos 61^\circ \sin 29^\circ\right]$$ is:

A

1

B

3

C

2

D

0

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Solution

As per the given question,

$$\left[\frac{\sin^2 24^\circ + \sin^2 66^\circ}{\cos^2 24^\circ + \cos^2 66^\circ} + \sin^2 61^\circ + \cos 61^\circ \sin 29^\circ\right]$$

We know that $$\sin(90^\circ-\theta=\cos \theta$$ and $$\cos(90^\circ-\theta)=\sin\theta$$, so this property in the above equation,

$$\Rightarrow \left[\frac{\sin^2 24^\circ + (\cos (90^\circ-66^\circ)^2}{\cos^2 24^\circ + (\sin (90^\circ-66^\circ)^2} + \sin^2 61^\circ + \cos 61^\circ \cos(90^\circ- 29^\circ)\right]$$

$$\Rightarrow \left[\frac{\sin^2 24^\circ + \cos^2 24^\circ}{\cos^2 24^\circ + \sin^2 24^\circ} + \sin^2 61^\circ + \cos 61^\circ \cos61^\circ\right]$$

We know that $$\sin^2\theta+\cos^2 \theta=1$$, so using this property in the above equation,

$$\Rightarrow \left[1 + \sin^2 61^\circ + \cos^2 61^\circ \right]$$

$$\Rightarrow 1 + 1=2$$

Question 149 Show answer

The value of $$\sin^2 32^\circ + \sin^2 58^\circ - \sin 30^\circ + \sec^2 60^\circ$$ is equal to:

A

5.5

B

3.5

C

4.5

D

4

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Solution

$$\sin \theta = \sin(90 - \theta) =\cos \theta$$ 

$$\sin^2 58^\circ = \cos^2 58^\circ$$

So , $$\sin^2 32^\circ + \cos^2 32^\circ - \sin 30^\circ + \sec^2 60^\circ$$

= 1 - $$\frac{1}{2}$$ + 4

=4.5

Question 150 Show answer

$$\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1}$$ = ?

A

$$\sec \theta + \tan \theta$$

B

$$\sec \theta \tan \theta$$

C

$$\sec \theta - \tan \theta$$

D

$$\sec \theta \sin \theta$$

View Explanation Hide Explanation

Solution

As per the given question,

$$\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1}$$

Now taking the conjugate of the  given equation,

$$\Rightarrow \frac{(\sin \theta - \cos \theta + 1)(\sin \theta + \cos \theta + 1)}{(\sin \theta + \cos \theta - 1)(\sin \theta + \cos \theta + 1)}$$

$$\Rightarrow \dfrac{\sin^2 \theta + \sin\theta \cos \theta +\sin \theta- \sin \theta \cos \theta-\cos^2\theta-\cos\theta+\sin\theta+\cos\theta+1}{\sin^2 \theta + \sin\theta \cos \theta +\sin \theta+ \sin \theta \cos \theta+\cos^2\theta+\cos\theta0-\sin\theta-\cos\theta-1}$$

Here few terms are in equal and opposite sign so they will cancel out to each other

$$\Rightarrow \dfrac{\sin^2\theta+2\sin\theta}{2\sin\theta\cos\theta}$$

$$\Rightarrow \dfrac{\sin\theta}{\cos\theta}+\dfrac{1}{\cos\theta}$$

$$\Rightarrow \tan\theta+\sec\theta$$

Question 151 Show answer

If $$3 \sin \theta = 4 \cos \theta,  then  \tan^2 \theta + \sin \theta - \cos \theta$$ is equal to:

A

$$\frac{88}{45}$$

B

2

C

$$\frac{89}{45}$$

D

$$\frac{17}{9}$$

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Solution

Given that,

$$3 \sin \theta = 4 \cos \theta$$

$$\tan \theta=\dfrac{4}{3}$$

Let AB=4 and BC=3

From the phythagoras theorem,

$$AB^2+BC^2=AC^2$$

Now, substituting the values,

$$\Rightarrow AC=\sqrt{AB^2+BC^2}$$

$$\Rightarrow AC=\sqrt{4^2+3^2}$$

$$\Rightarrow AC=\sqrt{25}=5$$

Now, substituting the values in  $$\tan^2 \theta + \sin \theta - \cos \theta$$

$$\Rightarrow \dfrac{4}{3}^2+\dfrac{4}{5}-\dfrac{3}{5}$$

$$\Rightarrow \dfrac{16}{9}+\dfrac{4-1}{5}$$

$$\Rightarrow \dfrac{16}{9}+\dfrac{1}{5}$$

$$\Rightarrow \dfrac{80}{45}+\dfrac{9}{45}$$

$$\Rightarrow \dfrac{89}{45}$$

Question 152 Show answer

If $$\sin \theta = 4 \cos \theta$$, then what is the value of $$\sin \theta \cos \theta$$ ?

A

$$\frac{2}{9}$$

B

$$\frac{3}{10}$$

C

$$\frac{4}{17}$$

D

$$\frac{3}{4}$$

View Explanation Hide Explanation

Solution

Given that,

$$\sin \theta = 4 \cos \theta$$

So,

$$\dfrac{\sin \theta}{ \cos \theta} = 4$$

$$\tan \theta = 4$$

$$\tan \theta=\dfrac{AB}{BC}=\dfrac{4}{1}$$

$$\Rightarrow AB^2+BC^2=AC^2$$

$$\Rightarrow 4^2+1= AC^2$$

$$\Rightarrow \sqrt{17} =AC$$

So

$$\sin \theta=\dfrac{AB}{AC}=\dfrac{4}{\sqrt{17}}$$

$$\cos \theta=\dfrac{BC}{AC}=\dfrac{1}{\sqrt{17}}$$

Now, substituting the values,

$$\sin \theta \cos \theta =\dfrac{4}{\sqrt{17}} \times \dfrac{1}{\sqrt{17}}=\dfrac{4}{17}$$

Question 153 Show answer

The value of $$\frac{1}{\sin \theta} - \frac{\cot^2 \theta}{1 + \cosec \theta}$$ is:

A

2

B

0

C

-1

D

1

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Solution

$$\frac{1}{\sin \theta} - \frac{\cot^2 \theta}{1 + \cosec \theta}$$

where θ is replaced by t

Using cot(t)=$$\frac{cost}{sint}$$ and cosec=$$\frac{1}{sint}$$

$$\frac{1}{sint}$$ - $$\frac{cot^2t}{1+cosect}$$

$$\frac{1}{sint}$$ - $$\frac{cos^2t÷sin^2t}{1+(1÷sint)}$$

$$\frac{1}{sint}$$ - $$\frac{cos^2t÷sin^2t}{(sint+1)÷sint}$$

$$\frac{1}{sint}$$ - $$\frac{cos^2t}{sin^2t}$$ × $$\frac{sint}{sint+1}$$

$$\frac{1}{sint}$$ - $$\frac{cos^2t}{sint(sint+1)}$$

use this identity  $$\cos^2 \theta +\sin^2 \theta$$=1

hence $$\frac{1}{sint}$$ - $$\frac{1-sin^2t}{sint(sint+1)}$$

Use this identity $$(a^2-b^2)$$= (a+b)(a-b)

=$$\frac{1}{sint}$$ - $$\frac{(1+sint) × (1-sint )}{sint(sint+1)}$$

=$$\frac{1}{sint}$$ - $$\frac{1-sint }{sint}$$

=$$\frac{1-(1-sint)}{sint}$$

=$$\frac{1-(1+sint)}{sint}$$

=$$\frac{sint}{sint}$$

= 1

Question 154 Show answer

The value of $$\sin^230^{\circ}\cos^245^{\circ}+4\tan^230^{\circ}+\frac{1}{2}\sin^290^{\circ}+2\cos90^{\circ}$$ is:

A

2

B

$$\frac{47}{24}$$

C

$$\frac{15}{8}$$

D

$$\frac{23}{12}$$

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Solution

$$\sin^230^{\circ}\cos^245^{\circ}+4\tan^230^{\circ}+\frac{1}{2}\sin^290^{\circ}+2\cos90^{\circ}=\left(\frac{1}{2}\right)^2\left(\frac{1}{\sqrt{2}}\right)^2+4\left(\frac{1}{\sqrt{3}}\right)^2+\frac{1}{2}\left(1\right)^2+2\left(0\right)$$

$$=\left(\frac{1}{4}\right)\left(\frac{1}{2}\right)+4\left(\frac{1}{3}\right)+\frac{1}{2}$$

$$=\frac{1}{8}+\frac{4}{3}+\frac{1}{2}$$

$$=\frac{3+32+12}{24}$$

$$=\frac{47}{24}$$

Hence, the correct answer is Option B

Question 155 Show answer

 If $$tan\theta-cot\theta=0$$  and  $$\theta$$ is  positive acute angle then the value of  $$ \frac{tan(\theta+15)}{tan(\theta-15)}$$

A

$$\frac{1}{3}$$

B

$$\sqrt{3}$$

C

$$\frac{1}{\sqrt{3}}$$

D

3

Question 156 Show answer

The value of $$\sin^2 38^\circ + \sin^2 52^\circ + \sin^2 30^\circ - \tan^2 45^\circ$$ is equal to:

A

$$\frac{1}{3}$$

B

$$\frac{1}{4}$$

C

$$\frac{3}{4}$$

D

$$\frac{1}{2}$$

View Explanation Hide Explanation

Solution

$$\sin^2 38^\circ + \sin^2 52^\circ + \sin^2 30^\circ - \tan^2 45^\circ$$

$$\sin^2 38^\circ +\sin^2 (90-38)^\circ + \sin^2 30^\circ - \tan^2 45^\circ$$

$$\sin^2 38^\circ +\cos^2 38^\circ + \sin^2 30^\circ - \tan^2 45^\circ$$

$$1+\frac{1}{4}-1=\frac{1}{4}$$

Question 157 Show answer

The value $$\cosec(67^\circ + \theta) - \sec(23^\circ - \theta) + \cos 15^\circ \cos 35^\circ \cos 55^\circ \cos 60^\circ \cos 75^\circ$$ is:

A

2

B

0

C

1

D

$$\frac{1}{2}$$

View Explanation Hide Explanation

Solution

$$\cosec(67^\circ + \theta) - \sec(23^\circ - \theta) + \cos 15^\circ \cos 35^\circ \cos 55^\circ \cos 60^\circ \cos 75^\circ$$
= $$\cosec(67^\circ + \theta) - \sec(90 - 67^\circ + \theta) + \cos 15^\circ \cos 35^\circ \cos (90 - 35^\circ) \cos 60^\circ \cos(90 - 15^\circ)$$
= $$\cosec(67^\circ + \theta) - \cosec(67^\circ + \theta) + \cos 15^\circ \cos 35^\circ \sec 35^\circ \cos 60^\circ \sec 15^\circ$$
= $$\cos 60^\circ$$
= $$\frac{1}{2}$$

Question 158 Show answer

If $$\sec \theta = \frac{13}{5},  then  \tan \theta - \sin \theta + \cos \theta$$ is equal to:

A

$$\frac{23}{13}$$

B

$$\frac{118}{65}$$

C

$$\frac{121}{65}$$

D

$$\frac{124}{65}$$

View Explanation Hide Explanation Video Solution

Solution

$$\sec\theta$$ = $$\frac{13}{5}$$

$$\cos\theta$$ = $$\frac{5}{13}$$

$$\sin\theta$$ =  $$\sqrt{1- \cos^2 \theta}$$ = $$\sqrt{1- \frac{5}{13}^{2}}$$ = $$\frac{12}{13}$$

$$\tan\theta$$ = $$\frac {sin\theta}{cos\theta}$$ = $$\frac{12}{13} \div \frac{5}{13}$$ = $$\frac{12}{5}$$

$$\tan \theta - \sin \theta + \cos \theta$$ = $$\frac{12}{5}$$ - $$\frac{12}{13}$$ + $$\frac{5}{13}$$ = $$\frac{121}{65}$$

Question 159 Show answer

The value of $$\cot^2 62^\circ - \sec^2 28^\circ + \cosec^2 30^\circ + \tan^2 60^\circ $$ is equal to:

A

$$\frac{16}{3}$$

B

8

C

$$\frac{10}{3}$$

D

6

Question 160 Show answer

The value of $$\frac{\sin (78^\circ + \theta) - \cos (12^\circ - \theta) + (\tan^2 70^\circ - \cosec^2 20^\circ)}{\sin 25^\circ \cos 65^\circ + \cos 25^\circ \sin 65^\circ}$$ is:

A

2

B

-1

C

-2

D

0

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Solution

$$\frac{\sin (78^\circ + \theta) - \cos (12^\circ - \theta) + (\tan^2 70^\circ - \cosec^2 20^\circ)}{\sin 25^\circ \cos 65^\circ + \cos 25^\circ \sin 65^\circ}$$
= $$\frac{\sin (78^\circ + \theta) - \cos (90 - 78^\circ + \theta) + (\tan^2 70^\circ - \cosec^2(90 - 70^\circ)}{\sin 25^\circ \cos(90 -  25^\circ) + \cos 25^\circ \sin (90 - 25^\circ)}$$
= $$\frac{\sin (78^\circ + \theta) - \sin (78^\circ + \theta) + (\tan^2 70^\circ - \sec^270^\circ)}{\sin 25^\circ \sin25^\circ + \cos 25^\circ \cos 25^\circ}$$
= $$ \frac{(\tan^2 70^\circ - \sec^270^\circ)}{\sin^2 25^\circ  + \cos^2 25^\circ}$$ = -1

Question 161 Show answer

The value of $$\sin^2 60^\circ - \cos^2 45^\circ + \sec 60^\circ + \cos^2 40^\circ + \cos^2 50^\circ$$ is equal to:

A

$$\frac{9}{14}$$

B

$$\frac{13}{4}$$

C

$$\frac{11}{4}$$

D

$$\frac{7}{2}$$

Question 162 Show answer

If $$\sin \theta = \frac{p^2 -1}{p^2 + 1}$$, then $$\cos \theta $$ is equal to:

A

$$\frac{2p}{1+p^2}$$

B

$$\frac{p}{p^2-1}$$

C

$$\frac{p}{1+p^2}$$

D

$$\frac{2p}{p^2-1}$$

View Explanation Hide Explanation

Solution

$$\sin \theta = \frac{p^2 -1}{p^2 + 1}$$

We know that $$\sin ^2 \theta + \cos ^2 \theta = 1$$

=> $$ cos ^2 \theta = 1 - \sin^2 \theta $$

=> $$ cos \theta = \sqrt {1-sin^2\theta}$$

$$\cos \theta = \sqrt{1- (\frac{p^2 -1}{p^2 + 1})^2}$$

= $$\sqrt{1- (\frac{p^4 +1 -2p^2}{p^4+ 1 +2p^2})}$$

=$$\sqrt{ \frac{p^4+ 1 +2p^2 - p^4 -1 + 2p^2}{p^4+ 1 +2p^2})}$$

=$$\sqrt{ \frac{ 4p^2}{p^4+ 1 +2p^2}}$$

=$$\frac{2p}{1+p^2}$$

So , the answer would be Option a)$$\frac{2p}{1+p^2}$$.

Question 163 Show answer

If $$\tan 4\theta = \cot(40^\circ - 2\theta), then  \theta$$ is equal to:

A

$$25^\circ$$

B

$$30^\circ$$

C

$$20^\circ$$

D

$$35^\circ$$

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Solution

As per the question,

$$\tan 4\theta = \cot(40^\circ - 2\theta)$$

We know that $$\cot \theta=\tan(90^\circ- \theta)$$

So, $$\tan 4\theta = \tan(90^\circ -40^\circ + 2\theta)$$

$$\tan 4\theta = \tan(50^\circ + 2\theta)$$

Hence, $$4\theta =50^\circ +2 \theta$$

$$\Rightarrow 2\theta =50^\circ $$

$$\Rightarrow \theta =25^\circ $$

Question 164 Show answer

If $$12 \cot^2 \theta  - 31 \cosec \theta + 32 = 0, 0^\circ < \theta < 90^\circ$$, then the value of $$\tan \theta$$ will be:

A

$$\frac{4}{5},  \frac{4}{3}$$

B

$$\frac{4}{3},  \frac{3\sqrt7}{7}$$

C

$$\frac{5}{4},  \frac{4}{3}$$

D

$$\frac{4}{5},  \frac{5\sqrt7}{7}$$

View Explanation Hide Explanation

Solution

As per the given question,

$$12 \cot^2 \theta - 31 \cosec \theta + 32 = 0, 0^\circ < \theta < 90^\circ$$

$$12(\cosec^2 \theta-1)-31 \cosec \theta+32=0$$

$$12 \cosec^2 \theta-31 \cosec \theta+20=0$$

Let $$\cosec \theta=x$$

$$12 x^2-31x+20=0$$

$$12x^2-15x-16x+20=0$$

$$3x(4x-5)-4(4x-5)=0$$

$$(4x-5)(3x-4)=0$$

So, $$x=\dfrac{5}{4}$$ or $$x=\dfrac{4}{3}$$

So, $$\sin \theta=\dfrac{4}{5}$$ or $$\sin \theta=\dfrac{3}{4}$$

Hence, $$\tan \theta=\dfrac{4}{3} or \tan \theta=\dfrac{3}{\sqrt 7}$$

$$\tan \theta=\dfrac{4}{3} or \tan \theta=\dfrac{3\sqrt 7}{7}$$

Question 165 Show answer

If $$\sin 5 \theta = \cos(50^\circ - 3\theta),  then  \theta$$ is equal to:

A

$$25^\circ$$

B

$$15^\circ$$

C

$$20^\circ$$

D

$$30^\circ$$

View Explanation Hide Explanation

Solution

$$\sin5\theta=\cos(50^\circ-3\theta)$$

$$\cos\left(90^{\circ\ }-5\theta\ \right)=\cos\left(50^{\circ\ }-3\theta\ \right)$$

$$90^{\circ\ }-5\theta\ =50^{\circ\ }-3\theta\ $$

$$\theta=20^{\circ\ }\ $$

Question 166 Show answer

$$\left(\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta}\right)^2 + 1, \theta$$ ≠ $$45^\circ$$ is equal to:

A

$$2 \tan^2 \theta$$

B

$$\cot^2 \theta$$

C

$$\cosec^2 \theta$$

D

$$\sec^2 \theta$$

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Solution

As per the given question,

$$\Rightarrow \left(\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta}\right)^2 + 1$$

Now, $$\Rightarrow \dfrac{\sin^2\theta}{\cos^2\theta} \left(\frac{1- 2 \sin^2 \theta}{2 \cos^2 \theta - 1}\right)^2 + 1$$

We know that $$\Rightarrow \cos 2\theta=2\cos^2\theta =1-2\sin^2\theta=\cos^2\theta-\sin^2\theta$$

Now, $$\Rightarrow \dfrac{\sin^2\theta}{\cos^2\theta} \left(\frac{\cos 2 \theta}{\cos2\theta}\right)^2 + 1$$

$$\Rightarrow \dfrac{\sin^2\theta}{\cos^2\theta}  + 1$$

Now taking LCM ,

$$\Rightarrow \dfrac{\sin^2\theta+\cos^2\theta}{\cos^2\theta}=\dfrac{1}{\cos^2\theta}=\sec^2\theta$$

Question 167 Show answer

The value of $$\frac{(\sin \theta - \cos \theta)(1 + \tan \theta + \cot \theta)}{1 + \sin \theta \cos \theta} = ?$$

A

$$\sec \theta - \cosec \theta$$

B

$$\cosec \theta - \sec \theta$$

C

$$\sin \theta + \cos \theta$$

D

$$\tan \theta - \cot \theta$$

View Explanation Hide Explanation

Solution

$$\frac{(\sin \theta - \cos \theta)(1 + \tan \theta + \cot \theta)}{1 + \sin \theta \cos \theta}$$
Let the $$\theta$$ be $$45\degree$$,
= $$\frac{(\sin 45\degree- \cos 45\degree)(1 + \tan 45\degree + \cot 45\degree)}{1 + \sin 45\degree \cos 45\degree}$$
=  $$\frac{(\frac{1}{\sqrt{2}}- \frac{1}{\sqrt{2}})(1 + 1 + 1)}{1 + \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}}$$
= $$\frac{(0)(1 + 1 + 1)}{1 + \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}}$$ = 0
From the option A,
$$\sec \theta - \cosec \theta$$
On put the $$\theta$$ = $$45\degree$$,
= $$\sec 45\degree - \cosec 45\degree$$
= \sqrt{2} - \sqrt{2} = 0

Question 168 Show answer

The value of $$\sqrt{\frac{\cosec \phi - \cot \phi}{\cosec \phi + \cot \phi}} \div \frac{\sin \phi}{1 + \cos \phi}$$ is equal to:

A

$$\cosec \phi$$

B

$$\frac{1}{2}$$

C

$$\sec \phi$$

D

$$1$$

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Solution

$$\sqrt{\frac{\cosec \phi - \cot \phi}{\cosec \phi + \cot \phi}} \div \frac{\sin \phi}{1 + \cos \phi}$$
= $$\sqrt{\frac{\cosec \phi - \cot \phi}{\cosec \phi + \cot \phi}} \times \frac{1 + \cos \phi}{\sin \phi}$$
= $$\sqrt{\frac{\cosec \phi - \cot \phi}{\cosec \phi + \cot \phi} \times \frac{\cosec \phi - \cot \phi}{\cosec \phi - \cot \phi}} \times \frac{1 + \cos \phi}{\sin \phi}$$
= $$\sqrt{\frac{(\cosec \phi - \cot \phi)^2}{\cosec^2 \phi - \cot^2 \phi}} \times \frac{1 + \cos \phi}{\sin \phi}$$
= $$(\cosec \phi - \cot \phi) \times \frac{1 + \cos \phi}{\sin \phi}$$
= $$(\cosec \phi - \cot \phi) \times (\cosec \phi + \cot \phi)$$
= $$\cosec^2 \phi - \cot^2 \phi$$ = 1

Question 169 Show answer

when $$2 \sin^2 \theta = 3 \cos \theta,  and  0 \leq \theta \leq 90^\circ,  then  \theta = ? $$

A

$$60^\circ$$

B

$$90^\circ$$

C

$$30^\circ$$

D

$$45^\circ$$

Question 170 Show answer

If $$12 \sin \theta = 5 \cos \theta,  then  \sin \theta + \cos \theta - \cot \theta$$ is equal to:

A

$$-\frac{71}{65}$$

B

$$\frac{139}{156}$$

C

$$\frac{116}{156}$$

D

$$-\frac{16}{65}$$

View Explanation Hide Explanation

Solution

We get $$\tan\ \theta\ =\frac{12}{5}$$
Now sin +cos -cot =12/13+5/13 -5/12
=139/156

Question 171 Show answer

The value of $$\sec^2 28^\circ - \cot^2 62^\circ + \sin^2 60^\circ + \cosec^2 30^\circ$$ is equal to:

A

$$\frac{19}{4}$$

B

3

C

$$\frac{7}{2}$$

D

$$\frac{23}{4}$$

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Solution

$$\sec^2 28^\circ - \cot^2 62^\circ + \sin^2 60^\circ + \cosec^2 30^\circ$$

We know that $$\sec(90^\circ -A)=\cosec A, \cosec(90^\circ -A)=\sec A$$ and $$\sin^2 A+\cos^2 A=1 $$

Now, $$\sec^2 28^\circ - \cot^2 62^\circ + \sin^2 60^\circ + \cosec^2 30^\circ$$

$$\Rightarrow \cosec^2 (90^\circ-28^\circ) - \cot^2 62^\circ + \sin^2 60^\circ + \sec^2 (90^\circ- 30^\circ$$

$$\Rightarrow \cosec^2 62^\circ - \cot^2 62^\circ + \sin^2 60^\circ + \sec^2 60^\circ$$

$$\Rightarrow \dfrac{1}{\sin^2 62^\circ}-\dfrac{\cos^2 62^\circ}{\sin^2 62^\circ} + \sin^2 60^\circ + \dfrac{1}{\cos^2 60^\circ}$$

$$\Rightarrow \dfrac{1-\cos^2 62^\circ}{\sin^2 62^\circ}+ \sin^2 60^\circ + \dfrac{1}{\cos^2 60}$$

$$\Rightarrow \dfrac{\sin^2 62^\circ}{\sin^2 62^\circ}+ \sin^2 60^\circ + \dfrac{1}{\cos^2 60}$$

$$\Rightarrow 1+(\dfrac{\sqrt 3}{2})^2 + (\dfrac{1}{\dfrac{1}{2}})^2$$

$$\Rightarrow 1+\dfrac{3}{4} + 4$$

$$\Rightarrow \dfrac{4+3+16}{4}$$

$$\Rightarrow \dfrac{23}{4}$$

Question 172 Show answer

$$\frac{2 + \tan^2 \theta + \cot^2 \theta}{\sec \theta \cosec \theta}$$ is equal to:

A

$$\cot \theta$$

B

$$\cos \theta \sin \theta$$

C

$$\sec \theta \cosec \theta$$

D

$$\tan \theta$$

View Explanation Hide Explanation

Solution

$$\frac{2 + \tan^2 \theta + \cot^2 \theta}{\sec \theta \cosec \theta}$$

= $$\frac{1 + \tan^2 \theta +1 + \cot^2 \theta}{\sec \theta \cosec \theta}$$

=$$\frac{ \sec^2 \theta + \cosec^2 \theta}{\sec \theta \cosec \theta}$$

=$$\frac{ \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}}{\frac{1}{sin\theta cos\theta}}$$

= $$\frac{1}{sin\theta cos\theta}$$

= $$\sec\theta\cosec\theta$$

So , the answer would be option c)$$\sec\theta\cosec\theta$$.

Question 173 Show answer

The value of: $$\frac{\sin 44^\circ}{\cos 46^\circ} + \sin^2 60^\circ - \cos^2 45^\circ + \sec 60^\circ$$ is equal to:

A

$$\frac{13}{4}$$

B

$$\frac{11}{3}$$

C

$$\frac{11}{4}$$

D

$$\frac{7}{4}$$

View Explanation Hide Explanation

Solution

Given that,

$$\frac{\sin 44^\circ}{\cos 46^\circ} + \sin^2 60^\circ - \cos^2 45^\circ + \sec 60^\circ$$

$$\sin(90^\circ-\theta)=\cos(\theta)$$

$$\Rightarrow \frac{\cos (90-46)^\circ}{\cos 46^\circ} + \sin^2 60^\circ - \cos^2 45^\circ + \sec 60^\circ$$

$$\Rightarrow \frac{\cos 46^\circ}{\cos 46^\circ} + \sin^2 60^\circ - \cos^2 45^\circ + \sec 60^\circ$$

$$\Rightarrow 1 + \sin^2 60^\circ - \cos^2 45^\circ + \sec 60^\circ$$

$$\Rightarrow 1 +\dfrac{3}{4} - \dfrac{1}{2} +2$$

$$\Rightarrow \dfrac{4+3-2+8}{4}=\dfrac{13}{4}$$

Question 174 Show answer

If $$3 \sin \theta = 2 \cos \theta,  then  \frac{4 \sin \theta - \cos \theta}{4 \cos \theta + \sin \theta}$$ is equal to:

A

$$\frac{5}{7}$$

B

$$\frac{5}{8}$$

C

$$\frac{5}{14}$$

D

$$\frac{5}{11}$$

View Explanation Hide Explanation

Solution

$$3 \sin \theta = 2 \cos \theta$$

$$\frac{\sin \theta}{\cos \theta} =\frac{2}{3}$$

$$ \frac{4 \sin \theta - \cos \theta}{4 \cos \theta + \sin \theta}$$

Divide both numerator and denominator by $$\cos\theta$$ ,

$$\frac{4\tan \theta -1 }{4 + \tan \theta}$$

= $$\frac{5}{14}$$

So , the answer would be option c)$$\frac{5}{14}$$.

Question 175 Show answer

If $$\cos \theta = \frac{4}{5},  then  \sin^2 \theta \cos \theta + \cos^2 \theta \sin \theta$$ is equal to:

A

$$\frac{16}{25}$$

B

$$\frac{84}{125}$$

C

$$\frac{14}{25}$$

D

$$\frac{82}{125}$$

View Explanation Hide Explanation

Solution

Given that,

$$\cos \theta = \frac{4}{5}$$

So, $$\sin \theta=\sqrt{1-\cos^2 \theta}=\sqrt{1-(\dfrac{4}{5})^2}=\dfrac{3}{5}$$

Now, substituting the values in   $$\sin^2 \theta \cos \theta + \cos^2 \theta \sin \theta$$

$$\Rightarrow (\dfrac{3}{5})^2\times \dfrac{4}{5} +(\dfrac{4}{5})^2\times \dfrac{3}{5}$$

$$\Rightarrow \dfrac{36}{125}+\dfrac{48}{125}$$

$$\Rightarrow \dfrac{84}{125}$$

Question 176 Show answer

If $$\cosec 2 \theta = \sec (3 \theta - 15^\circ),  then  \theta $$ is equal to:

A

$$22^\circ$$

B

$$20^\circ$$

C

$$25^\circ$$

D

$$21^\circ$$

View Explanation Hide Explanation

Solution

$$\cosec2\theta = \sec(3\theta - 15)$$

=$$ \sec(3\theta +90 - 105)$$

= $$\sec(90 - (-3\theta + 105)$$

{$$sec(90 - \theta) = cosec\theta$$}

$$\cosec2\theta = \cosec(-3\theta + 105)$$

$$2\theta = -3\theta + 105$$

$$\theta = 21$$

So , the answer would be option d)$$21^\circ$$.

Question 177 Show answer

If $$\cosec 3\theta = \sec(20^\circ + 2\theta),  then  \theta$$ is equal to:

A

$$14^\circ$$

B

$$20^\circ$$

C

$$30^\circ$$

D

$$15^\circ$$

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Solution

$$\cosec 3\theta = \sec(20^\circ + 2\theta)$$

We know that $$\cosec (90^\circ-\theta)=\sec \theta$$

$$\Rightarrow \cosec 3\theta = \cosec(90^\circ - 20^\circ - 2\theta)$$

$$\Rightarrow \cosec 3\theta = \cosec(70^\circ - 2\theta)$$

From the above,

$$\Rightarrow 3\theta=70^\circ-2\theta$$

$$\Rightarrow 5\theta=70^\circ$$

$$\Rightarrow \theta=14^\circ$$

Question 178 Show answer

The value of $$\frac{\sin \theta + \cos \theta - 1}{\sin \theta - \cos \theta + 1} \times \frac{\tan^2 \theta(\cosec^2 \theta -1)}{\sec \theta - \tan \theta}$$ is:

A

0

B

-1

C

1

D

$$\frac{1}{2}$$

View Explanation Hide Explanation

Solution

$$\frac{\sin \theta + \cos \theta - 1}{\sin \theta - \cos \theta + 1} \times \frac{\tan^2 \theta(\cosec^2 \theta -1)}{\sec \theta - \tan \theta}$$
Put the $$\theta = 30\degree$$,
= $$\frac{\sin 30\degree + \cos 30\degree - 1}{\sin 30\degree - \cos 30\degree + 1} \times \frac{\tan^2 30\degree(\cosec^2 30\degree -1)}{\sec 30\degree - \tan 30\degree}$$
= $$\frac{\frac{1}{2} + \frac{\sqrt{3}}{2} - 1}{\frac{1}{2} - \frac{\sqrt{3}}{2} + 1} \times \frac{\frac{1}{3}(4 -1)}{\frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}}}$$
= $$\frac{\frac{\sqrt{3}}{2} - \frac{1}{2}}{\frac{3}{2} - \frac{\sqrt{3}}{2}} \times \frac{1}{\frac{1}{\sqrt{3}}}$$
= $$\frac{\sqrt{3} - 1}{3 - \sqrt{3}} \times  \sqrt{3}$$
= $$\frac{3 - \sqrt{3}}{3 - \sqrt{3}}$$ = 1

Question 179 Show answer

If $$\tan \theta = \frac{3}{4},  then  \frac{4 \sin \theta - \cos \theta}{4 \sin \theta - \cos \theta}$$ is equal to:

A

$$\frac{3}{5}$$

B

$$\frac{1}{4}$$

C

$$\frac{1}{2}$$

D

$$\frac{2}{5}$$

Question 180 Show answer

Let $$a = \frac{2 \sin x}{1 + \sin x + \cos x}$$ and $$b = \frac{c}{1 + \sin x}$$. Then a = b, if c = ?

A

$$1 - \sin x \cos x$$

B

$$1 + \sin x - \cos x$$

C

$$1 + \sin x  \cos x$$

D

$$1 + \cos x - \sin x$$

View Explanation Hide Explanation

Solution

Given that,

$$a = \frac{2 \sin x}{1 + \sin x + \cos x}$$ and $$b = \frac{c}{1 + \sin x}$$

If $$a=b$$

$$ \frac{2 \sin x}{1 + \sin x + \cos x}=\frac{c}{1 + \sin x}$$

$$\Rightarrow c= \frac{2 \sin x(1 + \sin x)}{1 + \sin x + \cos x}$$

Now, multiplying and dividing by the conjugate of the denominator, in the given ratio

$$\Rightarrow c= \frac{2 \sin x(1 + \sin x)(1 + \sin x - \cos x)}{(1 + \sin x + \cos x)(1 + \sin x - \cos x)}$$

$$\Rightarrow c= \frac{2 \sin x(1 + \sin x)(1 + \sin x - \cos x)}{(1 + \sin x)^2 -( \cos x)^2}$$

$$\Rightarrow c= \frac{2 \sin x(1 + \sin x)(1 + \sin x - \cos x)}{(1 + \sin^2 x+2\sin x - \cos^2 x)}$$

$$\Rightarrow c= \frac{2 \sin x(1 + \sin x)(1 + \sin x - \cos x)}{(\sin^2 x+\cos^2 x + \sin^2 x+2\sin x - \cos^2 x)}$$

$$\Rightarrow c= \frac{2 \sin x(1 + \sin x)(1 + \sin x - \cos x)}{(2 \sin^2 x+ x+2\sin x) }$$

$$\Rightarrow c= \frac{2 \sin x(1 + \sin x)(1 + \sin x - \cos x)}{2\sin x( \sin x+1 )}$$

$$\Rightarrow c= 1 + \sin x - \cos x$$

Question 181 Show answer

The value of $$\frac{1}{\sec x - \tan x} - \frac{1}{\cos x}, 0^\circ < x < 90^\circ,$$ is equal to:

A

$$2 \cos x$$

B

$$\tan x$$

C

$$2 \sec x$$

D

$$\cot x$$

View Explanation Hide Explanation

Solution

$$\frac{1}{\sec x - \tan x} - \frac{1}{\cos x}$$

simplifying the sum secx-tanx

use the identity tanx=$$\frac{sinx}{cosx}$$

then use the identity secx=$$\frac{1}{cosx}$$

$$\frac{1}{1÷cosx-sinx÷cosx}$$ - $$\frac{1}{cosx}$$

$$\frac{1}{1-sinx÷cosx}$$ - $$\frac{1}{cosx}$$

$$\frac{cosx}{1-sinx}$$ - $$\frac{-1}{cosx}$$

order factor for (1-sinx)cosx

$$\frac{cos^2x-1+sinx}{cosx(1-sinx)}$$

use the identity 

$$\cos^2 \theta +\sin^2 \theta$$=1

$$\sin^2 \theta$$=1-$$\cos^2 \theta$$

-$$\sin^2 \theta$$ = -(1-$$\cos^2 \theta$$)

-$$\sin^2 \theta$$= -1+$$\cos^2 \theta$$

where θ is replaced by x

Hence =  $$\frac{-1+cos^2x+sinx}{cosx(1-sinx)}$$

           = $$\frac{-sin^2x+sinx}{cosx(1-sinx)}$$

.          = $$\frac{sinx(1-sinx)}{cosx(1-sinx)}$$

cancel (1-sinx)(1-sinx)

then =$$\frac{sinx}{cosx}$$

by using the identity we know that tanx=$$\frac{sinx}{cosx}$$

Question 182 Show answer

The value of $$\sqrt{\sec^2 \theta + \cosec^2 \theta} \times \sqrt{\tan^2 \theta - \sin^2 \theta}$$ is equal to:

A

$$\sin \theta \sec^2 \theta$$

B

$$\cosec \theta \sec^2 \theta$$

C

$$\cosec \theta \cos^2 \theta$$

D

$$\sin \theta \cos^2 \theta$$

View Explanation Hide Explanation

Solution

As per the question,

$$\sqrt{\sec^2 \theta + \cosec^2 \theta} \times \sqrt{\tan^2 \theta - \sin^2 \theta}$$

$$\Rightarrow \sqrt{\dfrac{1}{\cos^2 \theta} + \dfrac{1}{\sin^2 \theta}} \times \sqrt{\dfrac{\sin^2\theta}{\cos^2 \theta} - \sin^2 \theta}$$

$$\Rightarrow \sqrt{\dfrac{\sin^2 \theta + \cos^2 \theta }{ \cos^2 \theta \sin^2 \theta} }\times\sqrt{\dfrac{\sin^2\theta}{\cos^2 \theta} - \sin^2 \theta}$$

We know that,$$ \sin^2 \theta + \cos^2 \theta=1$$

$$\Rightarrow \sqrt{\dfrac{\sin^2\theta}{ \cos^2 \theta \sin^2 \theta (\cos^2 \theta)} -\dfrac{ \sin^2\theta}{ \cos^2 \theta \sin^2 \theta}}$$

$$\Rightarrow \sqrt{\dfrac{1}{ \cos^4 \theta} -\dfrac{ 1}{ \cos^2 \theta }}$$

$$\Rightarrow \sqrt{\dfrac{1-\cos^2\theta}{ \cos^4 \theta }}$$

$$\Rightarrow \sqrt{\dfrac{\sin^2\theta}{ \cos^4 \theta }}$$

$$\Rightarrow \sin \theta \sec^2 \theta $$

Question 183 Show answer

If $$3 \cos^2 A + 7 \sin^2 A = 4$$, then what is the value of $$\cot A$$, given that A is an acute angle?

A

1

B

$$\frac{\sqrt3}{2}$$

C

$$\frac{1}{\sqrt3}$$

D

$$\sqrt3$$

View Explanation Hide Explanation

Solution

As per the given question,

$$3 \cos^2 A + 7 \sin^2 A = 4$$

$$\Rightarrow 3 \cos^2 A + 3 \sin^2 A +4 \sin^2 A = 4$$

We know that $$\sin^2 \theta+\cos^2\theta=1$$

$$\Rightarrow 3 (\cos^2 A +  \sin^2 A )+4 \sin^2 A = 4$$

$$\Rightarrow 3 (1 )+4 \sin^2 A = 4$$

$$\Rightarrow 4 \sin^2 A = 1$$

$$\Rightarrow \sin A = \dfrac{1}{2}=\sin\dfrac{\pi}{6}$$

$$\Rightarrow A=\dfrac{\pi}{6}$$

Hence, $$\cot\theta=\cot \dfrac{\pi}{6}=\sqrt3$$

Question 184 Show answer

If $$3 \sin \theta  = 2 \cos^2 \theta, 0^\circ < \theta < 90^\circ$$, then the value of $$(\tan^2 \theta + \sec^2 \theta - \cosec^2 \theta)$$ is:

A

-2

B

$$-\frac{7}{3}$$

C

$$\frac{7}{3}$$

D

2

Question 185 Show answer

If $$\cot \theta = \frac{3}{4},  then  \sin \theta + \cos \theta - \tan \theta$$ is equal to:

A

$$\frac{1}{15}$$

B

$$-\frac{1}{20}$$

C

$$\frac{1}{20}$$

D

$$\frac{2}{15}$$

View Explanation Hide Explanation

Solution

Given that,

$$\cot \theta = \frac{3}{4}$$
Then $$\tan \theta=\frac{4}{3}$$

We know for the right angled triangle,
$$Hypotenuse^2=Base^2+Perpendicular^2$$
So, $$ =\sqrt{3^2+4^2}$$
$$ =\sqrt{25}$$
$$ =5$$
Hence, $$\sin\theta=\dfrac{4}{5}$$ and $$\cos\theta=\dfrac{3}{5}$$ and

$$\Rightarrow \sin \theta + \cos \theta - \tan \theta$$
$$\Rightarrow \dfrac{4}{5}+ \dfrac{3}{5} - \dfrac{4}{3}$$
$$\Rightarrow \dfrac{4\times3}{5\times3}+ \dfrac{3\times3}{5\times3} - \dfrac{4\times5}{3\times5}$$
$$\Rightarrow \dfrac{12+9-20}{5\times3}$$
$$\Rightarrow \dfrac{1}{15}$$

Question 186 Show answer

If $$\frac{\tan \theta + \sin \theta}{\tan \theta - \sin \theta} = \frac{k + 1}{k - 1}$$, then k = ?

A

$$\cosec \theta$$

B

$$\sec \theta$$

C

$$\cos \theta$$

D

$$\sin \theta$$

Question 187 Show answer

$$\sec^2 29^\circ - \cot^2 61^\circ + \sin^2 60^\circ + \cosec^2 30^\circ$$ is equal to:

A

$$\frac{23}{4}$$

B

$$\frac{11}{4}$$

C

$$\frac{19}{4}$$

D

$$\frac{15}{4}$$

View Explanation Hide Explanation

Solution

We know that $$\sec(90^\circ-\theta)$$ = $$\cosec\theta$$

$$\sec29^\circ=\cosec61^\circ$$

Now, $$\cosec^{2}61^\circ-\cot^{2}61^\circ+\sin^{2}60^\circ+\cosec^{2}30^\circ$$

    = $$1+\frac{3}{4}+\frac{4}{1}$$     Since,  ( $$\cosec^{2}\theta-\cot^{2}\theta=1$$)

   = $$\frac{23}{4}$$

Question 188 Show answer

The value of $$\theta$$, when $$\sqrt3 \cos \theta + \sin \theta = 1 (0^\circ \leq \theta \leq 90^\circ)$$, is:

A

$$90^\circ$$

B

$$30^\circ$$

C

$$60^\circ$$

D

$$0^\circ$$

View Explanation Hide Explanation

Solution

As per the given question,

$$\sqrt3 \cos \theta + \sin \theta = 1 (0^\circ \leq \theta \leq 90^\circ)$$

Dividing the 2 on both side of the equation,

$$\Rightarrow \dfrac{\sqrt3}{2}\times \cos \theta +\dfrac{1}{2}\times \sin \theta = \dfrac{1}{2}$$

Now, $$\cos 30 \cos \theta +\sin 30 \sin \theta = \cos 60$$

We know that $$\cos(\theta+\phi)=\cos \theta \cos \phi -\sin \phi \sin \theta$$

$$\Rightarrow \cos(\theta-30) = \cos 60$$

$$\Rightarrow \theta-30=60$$

$$\Rightarrow \theta=90^\circ$$

Hence $$\theta =90^\circ$$

Question 189 Show answer

If $$\cos^2 \theta  - 3 \cos \theta + 2 = \sin^2 \theta, 0^\circ < \theta < 90^\circ$$, then the value of $$2 \cosec \theta + 4 \cot \theta$$ is:

A

$$\frac{8\sqrt3}{3}$$

B

$$\frac{4\sqrt3}{4}$$

C

$$2\sqrt{3}$$

D

$$4\sqrt{3}$$

Question 190 Show answer

If $$\frac{\cos \theta}{1 - \sin \theta} + \frac{\cos \theta}{1 + \sin \theta} = 4, 0^\circ < \theta < 90^\circ$$, then the value of $$(\tan \theta + \cosec \theta)$$ is:

A

$$\frac{5\sqrt2}{2}$$

B

$$\frac{5\sqrt2}{3}$$

C

$$\frac{5\sqrt3}{3}$$

D

$$\frac{4\sqrt3}{3}$$

Question 191 Show answer

If $$\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + k$$, then k = 

A

$$\cot \theta + \sec \theta$$

B

$$\tan \theta + \sec \theta$$

C

$$\tan \theta  \cosec \theta$$

D

$$\cosec \theta  \sec \theta$$

View Explanation Hide Explanation

Solution

= $$\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta}$$

= $$\frac{\tan \theta}{1 - \1÷tan \theta} + \frac{\cot \theta}{1 - \tan \theta}

Question 192 Show answer

If $$\tan x = \cot(45^\circ+ 2x)$$, then what is value of x?

A

$$15^\circ$$

B

$$20^\circ$$

C

$$45^\circ$$

D

$$\frac{45^\circ}{2}$$

View Explanation Hide Explanation

Solution

Given,

$$\tan x = \cot(45^\circ+ 2x)$$

We know that, $$\tan(90^\circ-\theta)=\cot \theta$$

$$\Rightarrow \tan x = \tan(90^\circ-45^\circ- 2x)$$

$$\Rightarrow \tan x = \tan(45^\circ- 2x)$$

$$\Rightarrow x = 45^\circ- 2x$$

$$\Rightarrow 3x = 45^\circ$$

$$\Rightarrow x = 15^\circ$$

Question 193 Show answer

The value of $$\sin^2 20^\circ + \sin^2 70^\circ - \tan^2 45^\circ + \sec 60^\circ$$ is equal to:

A

2

B

1

C

2.5

D

3

View Explanation Hide Explanation

Solution

Given that,

$$\sin^2 20^\circ + \sin^2 70^\circ - \tan^2 45^\circ + \sec 60^\circ$$

We know that, $$\cos(90-\theta)=\sin \theta$$

$$\sin^2 20^\circ + \cos^2 (90^\circ-70^\circ) - \tan^2 45^\circ + \sec 60^\circ$$

$$\Rightarrow \sin^2 20^\circ + \cos^2 20^\circ - \tan^2 45^\circ + \sec 60^\circ$$

We know that $$\sin^2 \theta +\cos^2 \theta=1$$

So,

$$\Rightarrow 1 - \tan^2 45^\circ + \sec 60^\circ$$

We know that $$\tan 45=1$$ and $$\cos 60^\circ=\dfrac{1}{2}$$

Now, substituting the values

$$\Rightarrow= 1 - 1+ \dfrac{1}{\cos 60^\circ}$$

$$\Rightarrow= 1 - 1+ \dfrac{2}{1}$$

$$\Rightarrow= 2$$

Question 194 Show answer

A student got 24% marks in an exam and he failed by 56 marks. If he got 60% marks, then his marks are 70 more than the minimum passing marks. What is the maximum marks for the exam?

A

350

B

260

C

380

D

280

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Solution

Let maximum marks in the exam = $$100x$$ and minimum passing marks = $$y$$

=> Minimum passing marks = $$y=24x+56$$ --------------(i)

Also, $$y=60x-70$$ ----------------(ii)

Comparing both the equations, we get :

=> $$24x+56=60x-70$$

=> $$36x=70+56=126$$

=> $$x=\frac{126}{36}=3.5$$

$$\therefore$$ Maximum marks = $$100\times3.5=350$$

=> Ans - (A)

Question 195 Show answer

If $$\left(\frac{\tan \theta - \sec \theta + 1}{\tan \theta + \sec \theta - 1}\right) \sec \theta = \frac{1}{k}$$, then k = 

A

$$1 + \sin \theta$$

B

$$1 - \cos \theta$$

C

$$1 + \cos \theta$$

D

$$1 - \sin \theta$$

Question 196 Show answer

If $$\sin \theta = \cos(50^\circ + \theta)$$, then $$\theta$$ is equal to:

A

$$20^\circ$$

B

$$35^\circ$$

C

$$25^\circ$$

D

$$30^\circ$$

View Explanation Hide Explanation

Solution

$$\sin \theta = \cos(50^\circ + \theta)$$

$$\cos\left(90^{\circ\ }-\theta\ \right)=\cos\left(50^{\circ\ }+\theta\ \right)$$

$$90^{\circ\ }-\theta\ =50^{\circ\ }+\theta\ $$

$$2\theta\ =40^{\circ\ }$$

$$\theta\ =20^{\circ\ }$$

Question 197 Show answer

If $$0^\circ < \theta < 90^\circ$$ and $$\cos^2 \theta = 3(\cot^2 \theta - \cos^2 \theta)$$ then the value of $$\left(\frac{1}{2} \sec \theta + \sin \theta \right)^{-1}$$ is:

A

$$2(\sqrt3 - 1)$$

B

$$\sqrt3 + 1$$

C

$$\sqrt3 + 2$$

D

$$2(2 - \sqrt3)$$

View Explanation Hide Explanation

Solution

From the given question,

$$\cos^2 \theta = 3(\cot^2 \theta - \cos^2 \theta)$$

$$\Rightarrow \cos^2 \theta = 3\left( \dfrac{\cos^2 \theta}{\sin^2 \theta} - \cos^2 \theta \right)$$

$$\Rightarrow \cos^2 \theta = 3\left( \dfrac{\cos^2 \theta- \cos^2 \theta \sin^2 \theta}{\sin^2 \theta} \right)$$

$$\Rightarrow \cos^2 \theta = 3 \cos^2 \theta\left( \dfrac{1- \sin^2 \theta}{\sin^2 \theta} \right)$$

$$\Rightarrow \cos^2 \theta = 3 \cos^2 \theta\left( \dfrac{ \cos^2 \theta}{\sin^2 \theta} \right)$$

$$\Rightarrow \cos^2 \theta \sin^2 \theta= 3 \cos^2 \theta\left( \cos^2 \theta \right)$$

$$\Rightarrow \cos^2 \theta \sin^2 \theta= 3 \cos^4 \theta$$

$$\Rightarrow 3 \cos^4 \theta -\cos^2 \theta \sin^2 \theta= 0$$

$$\Rightarrow 3 \cos^4 \theta -\cos^2 \theta(1- \cos^2 \theta)= 0$$

$$\Rightarrow 3 \cos^4 \theta -\cos^2 \theta+ \cos^4 \theta= 0$$

$$\Rightarrow 4 \cos^4 \theta -\cos^2 \theta= 0$$

$$\Rightarrow \cos^2(4 \cos^2 \theta -1)= 0$$

$$\Rightarrow 4 \cos^2 \theta -1= 0 $$ and $$\cos^2 \theta= 0$$

$$\Rightarrow 4 \cos^2 \theta= 1 $$ and $$\cos^2 \theta= 0$$

$$\Rightarrow \cos^2 \theta= \dfrac{1}{4} $$ and $$\cos \theta=\cos( \dfrac{ \pi}{2})= 0$$

$$\Rightarrow \cos \theta= \dfrac{1}{2}=\cos{\dfrac{\pi}{3}} $$ and $$\cos \theta=\cos( \dfrac{ \pi}{2})= 0$$

Hence, $$\theta=\dfrac{\pi}{3}$$ And $$\theta=\dfrac{\pi}{2}$$

As per the question it is given that $$ 0<\theta<\dfrac{\pi}{2}$$

Now, substituting the values of $$\theta=\dfrac{\pi}{3}$$

$$\Rightarrow \left(\frac{1}{2} \sec (\dfrac{\pi}{3}) + \sin (\dfrac{\pi}{3}) \right)^{-1}$$

$$\Rightarrow \left(\frac{1}{2} 2+ \dfrac{\sqrt 3}{2} \right)^{-1}$$

$$\Rightarrow \left(1+ \dfrac{\sqrt 3}{2} \right)^{-1}$$

$$\Rightarrow \left( \dfrac{2+\sqrt 3}{2} \right)^{-1}$$

$$\Rightarrow \left( \dfrac{2}{2+\sqrt 3} \right)$$

Now, taking the conjugate of the denominator,

$$\Rightarrow \left( \dfrac{2(2-\sqrt 3)}{(2+\sqrt 3)(2-\sqrt 3)}\right)= \dfrac{2(2-\sqrt 3)}{(4- 3)}= 2(2-\sqrt 3)$$

Question 198 Show answer

If $$\cosec 4\theta = \sec (60^\circ - 2\theta),  then  \theta $$ is equal to:

A

$$25^\circ$$

B

$$20^\circ$$

C

$$15^\circ$$

D

$$18^\circ$$

View Explanation Hide Explanation

Solution

since ,  $$cosecA=secB$$ if $$A+B$$ = $$90^\circ$$

 So, $$4\theta+(60^\circ-2\theta)$$ = $$90^\circ$$

    $$\Rightarrow$$ $$2\theta=30^\circ$$

     $$ \Rightarrow$$ $$\theta=15^\circ$$

Question 199 Show answer

If $$\sin^2 \alpha + \sin^2 \beta$$ , then the value of $$\cos \frac{\alpha + \beta}{2}$$ is

A

1

B

-1

C

0

D

0.5

Question 200 Show answer

The value of $$\cot \frac{\pi}{20} \cot \frac{3\pi}{20}\cot \frac{5\pi}{20}\cot \frac{7\pi}{20}\cot \frac{9\pi}{20}$$ is

A

-1

B

$$\frac{1}{2}$$

C

0

D

1

Question 201 Show answer

If $$\sin \theta + \cos \theta = \frac{17}{13} , 0 < \theta < 90^\circ$$ ,then the value of $$\sin \theta - \cos \theta$$ is

A

$$\frac{5}{17}$$

B

$$\frac{3}{19}$$

C

$$\frac{7}{10}$$

D

$$\frac{7}{13}$$

Question 202 Show answer

From the top of a tower, the angles of depression of two objects on the ground on the same side of it, are observed to be $$60^\circ$$ and $$30^\circ$$ respectively and the distance between the objects is $$400 \surd3$$ m. The height (in m) of the tower is:

A

800

B

800 $$\sqrt3$$

C

600

D

600 $$\sqrt3$$

View Explanation Hide Explanation

Solution

BC = $$400 \surd3$$ m
In $$\triangle$$ ACD,
$$tan 60\degree = \frac{\sqrt{3}}{1} = \frac{AD}{CD}$$ 
In $$\triangle$$ ABD,
$$tan 30\degree = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} = \frac{AD}{BD}$$
BC = BD - CD = 3 - 1 = 2 units
2 units =  $$400 \surd3$$
$$\sqrt{3} units = \frac{400 \surd3}{2} \times \sqrt{3}  = 600 m
The height of the tower = 600 m.

Question 203 Show answer

If $$\tan \theta .\tan 2\theta = 1$$, then the value of $$\sin^2 2\theta + \tan^2 2\theta$$ is equal to

A

$$\frac{3}{4}$$

B

$$\frac{10}{3}$$

C

$$3\frac{3}{4}$$

D

3

Question 204 Show answer

P and Q are two points on the ground on either side of a pole. The angles of elevation of the top of the pole as observed from P and Q are $$60^\circ$$ and $$30^\circ$$, respectively and the distance between them is $$84\sqrt3$$ m. What is the height (in m) of the pole?

A

63

B

73.5

C

52.5

D

60

View Explanation Hide Explanation

Solution

In $$\triangle POC$$,
$$\tan 60$$ = $$\frac{\sqrt{3}}{1} = \frac{OC}{OP}$$
$$\frac{OC}{OP} = \frac{3}{\sqrt{3}}$$
In $$\triangle QOC$$,
$$\tan 30$$ = $$\frac{1}{\sqrt{3}} = \frac{OC}{OQ}$$
$$\frac{OC}{OQ} = \frac{3}{3\sqrt{3}}$$
OC = 3 units
PQ = OP + OQ = $$\sqrt{3} + 3\sqrt{3} = 4\sqrt{3}$$ units
$$84\sqrt3 = 4\sqrt{3}$$ units
1 unit = 21 m
3 unit = 63 m
Height of the pole = 63 m

Question 205 Show answer

- If x tan $$60^\circ$$ + cos $$45^\circ$$ = sec $$45^\circ$$ then the value of $$x^{2}$$ + 1 is

A

$$\frac{6}{7}$$

B

$$\frac{7}{6}$$

C

$$\frac{5}{6}$$

D

$$\frac{6}{5}$$

View Explanation Hide Explanation

Solution

Given that x tan $$60^\circ$$ + cos $$45^\circ$$ = sec $$45^\circ$$

We know that value of tan $$60^\circ$$ = $$ \sqrt{3}$$, cos $$45^\circ$$ = $$\frac {1}{\sqrt{2}}$$ and sec $$45^\circ$$ = $$\sqrt{2}$$

=> x$$ \sqrt{3}$$+ $$\frac {1}{\sqrt{2}}$$ = $$\sqrt{2}$$

=> x$$ \sqrt{6}$$ + 1= 2

=> x$$ \sqrt{6}$$ = 1

=> x= $$ \frac{1}{\sqrt{6}}$$

Therefore $$ x^2 + 1 $$ = $$\frac{1} {6} $$ + 1 = $$ \frac{7}{6} $$

Question 206 Show answer

If $$5 cos\theta+ 12 sin\theta$$=13, $$0<\theta<90^\circ$$ then value of $$sin\theta$$

A

-$$\frac{12}{13}$$

B

$$\frac{5}{13}$$

C

$$\frac{12}{13}$$

D

$$\frac{6}{13}$$

Question 207 Show answer

If $$\sec A + \tan A = a$$, then the value of $$\cos A$$ is

A

$$\frac{a^2 + 1}{2a}$$

B

$$\frac{2a}{a^2 + 1}$$

C

$$\frac{a^2 - 1}{2a}$$

D

$$\frac{2a}{a^2 - 1}$$

Question 208 Show answer

The value of the expression $$(1 + \sec 22^\circ + \cot 68^\circ)(1 - \cosec 22^\circ + \tan 68^\circ)$$is

A

0

B

1

C

-1

D

2

View Explanation Hide Explanation

Solution

$$(1 + \sec 22^\circ + \cot 68^\circ)(1 - \cosec 22^\circ + \tan 68^\circ)$$

$$(1 + \sec 22^\circ + \tan22^\circ)(1 - \cosec 22^\circ + \cot22^\circ)$$

$$(1 + \frac{1}{cos 22^\circ} +  \frac{sin 22^\circ}{cos 22^\circ})(1 - \frac{1}{sin 22^\circ} +\frac{cos 22^\circ}{sin 22^\circ})$$

$$( \frac{1+cos 22^\circ +sin 22^\circ}{cos 22^\circ}) $$ $$( \frac{cos 22^\circ +sin 22^\circ -1}{sin 22^\circ})$$

$$\frac{sin^2 22^\circ + cos^2 22^\circ + 2cos22^\circ \times sin22^\circ -1}{ cos 22^\circ sin 22^\circ }$$

$$\frac{1 + 2cos 22^\circ sin 22^\circ -1}{ cos 22^\circ sin 22^\circ }$$ = 2

Question 209 Show answer

If $$\frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = \frac{4}{\sqrt{3}}, 0^\circ < \theta < 90^\circ$$, then the value of $$(\tan \theta + \sec \theta)^{-1}$$ is:

A

$$2 - \sqrt{3}$$

B

$$3 - \sqrt{2}$$

C

$$2 + \sqrt{3}$$

D

$$3 + \sqrt{2}$$

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Solution

$$\frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = \frac{4}{\sqrt{3}}$$
Put the $$\theta = 60\degree$$,
$$\frac{\sin 60\degree}{1 + \cos 60\degree} + \frac{1 + \cos 60\degree}{\sin 60\degree} = \frac{4}{\sqrt{3}}$$
$$\frac{\frac{\sqrt{3}}{2}}{1 +\frac{1}{2}} + \frac{1 + \frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{4}{\sqrt{3}}$$
$$\frac{\frac{3}{4} + \frac{9}{4}}{\frac{3}{2} \times \frac{3}{4}} = \frac{4}{\sqrt{3}}$$
$$\frac{4}{\sqrt{3}} = \frac{4}{\sqrt{3}}$$
Now,
$$(\tan \theta + \sec \theta)^{-1}$$
Put the $$\theta = 60\degree$$,
= $$(\tan 60\degree + \sec 60\degree)^{-1}$$
= $$(\sqrt{3} + 2)^{-1}$$
= $$2 - \sqrt{3}$$

Question 210 Show answer

If $$\sin P + \cosec P = 2$$, then the value of $$\sin^7 P + \cosec^7 P$$ is

A

1

B

2

C

3

D

0

Question 211 Show answer

If $$x \sin^3 \theta + y \cos^3 \theta = \sin \theta \cos \theta  and  x \sin \theta - y \cos \theta = 0$$, then the value of $$x^2 + y^2$$ equals

A

1

B

$$\frac{1}{2}$$

C

$$\frac{3}{2}$$

D

2

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Solution

$$x \sin^3 \theta + y \cos^3 \theta = \sin \theta \cos \theta $$ --> eq 1

$$ x \sin \theta - y \cos \theta = 0$$

$$ x \sin \theta = y \cos \theta$$ --->eq2

substituting in eq1

$$y\cos \theta \sin^2 \theta + y \cos^3 \theta = \sin \theta \cos \theta $$

taking $$ y\cos \theta  common$$

$$y\cos \theta (\sin^2 \theta +  \cos^2 \theta )= \sin \theta \cos \theta $$ { we know $$\sin^2 \theta +  \cos^2 \theta  = 1$$}

$$y\cos \theta= \sin \theta \cos \theta $$ 

$$y= \sin \theta$$

substituting in eq 2 

$$ x \sin \theta = \sin \theta \cos \theta$$ 

x = $$\cos \theta$$

 $$x^2 + y^2$$ =  $$\sin^2\theta + \cos^2\theta$$ = 1

Question 212 Show answer

x, y be two acute angles, x + y < $$90^\circ$$ and sin(2x -$$20^\circ$$) = cos(2y + $$20^\circ$$), the value of tan(x + y) is

A

$$\surd{3}$$

B

$$\frac{1}{\surd{3}}$$

C

1

D

2+$$\surd{2}$$

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Solution

The value of $$\sin\theta and \cos \theta are equal when \theta=45\degree$$

2x - 20 = 45 =>x = 32.5

2y + 20 =45 => y = 12.5

$$\tan(32.5+12.5) = \tan45\degree = 1$$

So, the answer would be option c)1

Question 213 Show answer

If $$a^{2}sec^{2} x-b^{2} tan^{2} x$$=$$c^{2}$$ then the value of $$sec^{2} x+tan^{2} x $$ is equal to ($$ b^{2} \neq  a^{2}$$)

A

$$\frac{b^{2}-a^{2}+2C^{2}}{b^{2}+a^{2}}$$

B

$$\frac{b^{2}+a^{2}-2C^{2}}{b^{2}-a^{2}}$$

C

$$\frac{b^{2}-a^{2}-2C^{2}}{b^{2}+a^{2}}$$

D

$$\frac{b^{2}-a^{2}}{b^{2}+a^{2}+2 C^{2}}$$

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Solution

Given that $$a^{2}sec^{2} x-b^{2} tan^{2} x$$=$$c^{2}$$

=> $$a^{2}(1+tan^{2} x)-b^{2} tan^{2} x$$=$$c^{2}$$

=> $$a^2+tan^{2}x(a^2-b^2)=c^{2}$$

=> $$tan^{2}x=\frac{c^2-a^2}{a^2-b^2}$$

Therefore, 

$$sec^{2} x+tan^{2} x $$ 

= $$1+2tan^{2}x$$

=$$1+2(\frac{c^2-a^2}{a^2-b^2})$$

=$$\frac{b^{2}+a^{2}-2C^{2}}{b^{2}-a^{2}}$$

Question 214 Show answer

If $$\cos x . \cos y + \sin x . \sin y = -1$$ then $$\cos x + \cos y$$ is

A

-2

B

1

C

0

D

2

Question 215 Show answer

If $$\sec \theta + \tan \theta = m (>1)$$, then the value of $$\sin \theta  is  (0^\circ < \theta < 90^\circ)$$

A

$$\frac{1 - m^2}{1 + m^2}$$

B

$$\frac{m^2 - 1}{m^2 + 1}$$

C

$$\frac{m^2 + 1}{m^2 - 1}$$

D

$$\frac{1 + m^2}{1 - m^2}$$

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Solution

 $$\sec \theta + \tan \theta = m (>1)$$

let $$\theta = 45\degree$$

$$\surd{2}+ 1 = m$$

$$m^2 = 3+2\surd{2}$$

$$m^2 - 1 = 3+2\surd{2} -1$$ = $$2+2\surd{2}$$

$$m^2 + 1 = 3+2\surd{2} +1$$ = $$2+2\surd{2}$$

$$\frac{m^2 - 1}{m^2 + 1} = \frac{2+2\surd{2}}{2+2\surd{2}}$$ = $$\frac{1}{\surd{2}} = sin\theta$$ 

 

Question 216 Show answer

-(1 + sec $$20^\circ$$ + cot $$70^\circ$$)(1 - cosec $$20^\circ$$ + tan$$70^\circ$$) is equal to

A

0

B

1

C

2

D

3

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Solution

(1 + sec $$20^\circ$$ + cot $$70^\circ$$)(1 - cosec $$20^\circ$$ + tan$$70^\circ$$)

= (1 + sec $$20^\circ$$ + tan $$20^\circ$$)(1 - cosec $$20^\circ$$ + cot$$20^\circ$$)

= (1 + $$\frac{1}{cos20^\circ}$$ + $$\frac{sin20^\circ}{cos20^\circ}$$)(1 - $$\frac{1}{sin20^\circ}$$ + $$\frac{cos20^\circ}{sin20^\circ}$$)

= ($$\frac{1+cos20^\circ+sin20^\circ}{cos20^\circ}$$)($$\frac{sin20^\circ-1+cos20^\circ}{sin20^\circ}$$)

=  $$\frac{(cos20^\circ+sin20^\circ)^2-1}{cos20^\circ sin20^\circ}$$

= $$\frac{2cos20^\circ sin20^\circ}{cos20^\circ sin20^\circ}$$

=2

Question 217 Show answer

If $$(a^2 - b^2)\sin \theta + 2ab \cos \theta = a^2 + b^2$$, then $$\tan \theta =$$

A

$$\frac{2ab}{a^2 - b^2}$$

B

$$\frac{a^2 - b^2}{2ab}$$

C

$$\frac{ab}{a^2 - b^2}$$

D

$$\frac{a^2 - b^2}{ab}$$

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Solution

$$(a^2 - b^2)\sin \theta + 2ab \cos \theta = a^2 + b^2$$

divide it by $$ a^2 + b^2$$

we get

$$\frac{(a^2 - b^2)\sin \theta}{a^2 + b^2} + \frac{2ab \cos \theta}{a^2 + b^2} = 1$$ ($$\because \sin^2 \theta + \cos^2 \theta = 1$$)

here $$\sin \theta = \frac{(a^2 - b^2)}{a^2 + b^2} $$

$$\cos \theta = \frac{2ab}{a^2 + b^2} $$

$$\tan \theta = \frac{\sin \theta}{\cos \theta}$$= $$ \frac{(a^2 - b^2)}{2ab}$$

Question 218 Show answer

The value of the expression $$2(\sin^6 \theta + \cos^6 \theta) -3 (\sin^4 \theta + \cos^4 \theta)+1$$ is

A

-1

B

0

C

1

D

2

Question 219 Show answer

If $$\cos \theta = \frac{x^2 - y^2}{x^2 + y^2}$$ then the value of $$\cot \theta$$ is equal to [If $$0 \leq \theta \leq 90^\circ$$]

A

$$\frac{2xy}{x^2 - y^2}$$

B

$$\frac{2xy}{x^2 + y^2}$$

C

$$\frac{x^2 + y^2}{2xy}$$

D

$$\frac{x^2 - y^2}{2xy}$$

Question 220 Show answer

 If $$tan ^4\theta + tan^2\theta$$ = 1 then the value of $$cos^4\theta + cos^2\theta$$ is

A

2

B

0

C

1

D

-1

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Solution

Is it $$tan 4\theta or \tan^4\theta$$

Question 221 Show answer

If $$2y \cos \theta = x \sin \theta  and  2x \sec \theta - y \cosec \theta = 3$$, then the value of $$x^2 + 4y^2$$ is

A

1

B

2

C

3

D

4

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Solution

$$let \theta  = 45\degree$$

$$2y \frac{1}{\sqrt{2}} = x\frac{1}{\sqrt{2}} $$ = 2y= x

$$ 2x \sqrt{2} - y \sqrt{2} = 3$$

2x - y = $$\frac{3}{\sqrt{2}}$$ {substituting y= $$\frac{x}{2}$$}

x= $$ \sqrt{2}$$

y= $$\frac{1}{ \sqrt{2}}$$

value of $$x^2 + 4y^2$$ = $$\sqrt{2}^2 + 4\frac{1}{ \sqrt{2}^2}$$ = 2+2 = 4

Question 222 Show answer

The distance between two pillars is 120 metres. The height of one pillar is thrice the other. The angles of elevation of their tops from the midpoint of the line connecting their feet are complementary to each other. The height (in metres) of the taller pillar is (Use : $$\surd3 = 1.732$$)

A

34.64

B

51.96

C

69.28

D

103.92

Question 223 Show answer

The value of 8$$(sin6\theta+cos6\theta)-(sin4\theta+cos4\theta)$$ is equal to

A

20

B

-20

C

-4

D

4

Question 224 Show answer

The value of (coseca — sina) (seca — cosa) (tana + cota)

A

4

B

2

C

1

D

6

Question 225 Show answer

If $$x = \cosec \theta - \sin \theta$$ and $$y = \sec \theta - \cos \theta$$, then the relation between x and y is

A

$$x^2 + y^2 + 3 = 1$$

B

$$x^2y^2(x^2 + y^2 + 3) = 1$$

C

$$x^2(x^2 + y^2 - 5) = 1$$

D

$$y^2(x^2 + y^2 - 5) = 1$$

Question 226 Show answer

A hydrogen filled balloon ascending at the rate of 18 kmph was drifted by wind. Its angle of elevation at 10th and 15th minutes were found to be $$60^\circ$$ and $$45^\circ$$ respectively. The wind speed (in whole numbers) during the last five minutes, approximately, is equal to

A

7 km/hr

B

11 km/hr

C

26 km/hr

D

33 km/hr

Question 227 Show answer

A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height h. At a point on the plane, the angle of elevation of the bottom of the flag staff is $$\alpha$$ and that of the top of the flag staff is $$\beta$$. Then the height of the tower is

A

$$h \tan \alpha$$

B

$$\frac{h \tan \alpha}{\tan \beta - \tan \alpha}$$

C

$$\frac{h \tan \alpha}{\tan \alpha - \tan \beta}$$

D

None of these

Question 228 Show answer

From a point exactly midway between the foot of two towers P and Q,the angles of elevation of their tops are $$30^\circ$$ and $$60^\circ$$, respectively. The ratio of the height of P to that of Q is:

A

1 : 3

B

1 : 2

C

$$1 : 2\sqrt{3}$$

D

$$2 : 3\sqrt{3}$$

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Solution

AO = OB
In $$\triangle AOP$$,
$$tan30\degree = \frac{AP}{AO}$$
$$\frac{1}{\sqrt{3}} = \frac{AP}{AO}$$
AP = 1
AO = $$\sqrt{3}$$

In $$\triangle AOP$$,
$$tan60\degree = \frac{BQ}{BO}$$
$$\frac{\sqrt{3}}{1} = \frac{BQ}{BO}$$
$$\frac{BQ}{BO} = \frac{3}{\sqrt{3}}$$
BQ = 3
BO = $$\sqrt{3}$$
The ratio of the height of P to that of Q = 1 : 3

Question 229 Show answer

The angle of elevation of an aeroplane as observed from a point 30 m above the transparent water-surface of a lake is $$30^\circ$$ and the angle of depression of the image of the aeroplane in the water of the lake is $$60^\circ$$. The height of the aeroplane from the water-surface of the lake is

A

60 m

B

45 m

C

50 m

D

75 m

Question 230 Show answer

The angles of depression of two ships from the top of a light house are $$60^\circ$$ and $$45^\circ$$ towards east. If the ships are 300 m apart, the height of the light house is

A

$$200(3 + \surd3)$$ meter

B

$$250(3 + \surd3)$$ meter

C

$$150(3 + \surd3)$$ meter

D

$$160(3 + \surd3)$$ meter

Question 231 Show answer

The expression of $$\frac{\cot \theta + \cosec \theta - 1}{\cot \theta + \cosec \theta + 1}$$ is equal to

A

$$\frac{1 + \cos \theta}{\sin \theta}$$

B

$$\frac{1 - \cos \theta}{\sin \theta}$$

C

$$\frac{\cot \theta + 1}{\cosec \theta}$$

D

$$\frac{\cot \theta - 1}{\cosec \theta}$$

Question 232 Show answer

The value of 

$$(\tan 29^\circ \cot 61^\circ - \cosec^2 61^\circ) + \cot^2 54^\circ - sec^2 36^\circ + (sin^2 1^\circ + sin^23^\circ + \sin^2 5^\circ + --- + \sin^2 89^\circ)$$ is:

A

$$20\frac{1}{2}$$

B

$$21$$

C

$$22\frac{1}{2}$$

D

$$22$$

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Solution

$$(\tan 29^\circ \cot 61^\circ - \cosec^2 61^\circ) + \cot^2 54^\circ - sec^2 36^\circ + (sin^2 1^\circ + sin^23^\circ + \sin^2 5^\circ + --- + \sin^2 89^\circ)$$
= $$(\tan (90 - 61^\circ) \cot 61^\circ - \cosec^2 61^\circ) + \cot^2 (90 - 36^\circ) - sec^2 36^\circ + (22 + sin^2 45^\circ$$
= $$(\cot^2 61^\circ - \cosec^2 61^\circ) + \cot^2  36^\circ- sec^2 36^\circ + 22 + \frac{1}{2}$$
= -1 - 1 + 22 + $$\frac{1}{2}$$
= $$20 \frac{1}{2}$$