Trigonometry Questions for SSC CHSL and MTS

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Question 1: 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) $\frac{3}{2}$

1) Answer (B)

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 2: 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) $\frac{23}{12}$

2) Answer (A)

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 3: 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) $\sqrt[4]{2}$

3) Answer (A)

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 4: 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) $45^\circ$

4) Answer (B)

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 5: 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) $-\frac{3}{2}$

5) Answer (B)

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

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Question 6: 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) $-2$

6) Answer (D)

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 7: 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) $3(2 + \sqrt{3})$

7) Answer (A)

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 8: 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

8) Answer (C)

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 9: 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

9) Answer (C)

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 10: 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

10) Answer (B)

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 11: 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) $\frac{11}{12}$

11) Answer (B)

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

a) -1

b) 0

c) 1

d) 2

12) Answer (C)

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 13: 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

13) Answer (A)

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 14: 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) $3\sqrt{3}$

14) Answer (B)

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

a) $\cos A$

b) $\sin A$

c) $\cot A$

d) $\cosec A$

15) Answer (B)

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 16: 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) $\sec^2 \theta$

16) Answer (B)

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 17: 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

17) Answer (C)

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: 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

18) Answer (B)

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 19: 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) $-\frac{19}{8}$

19) Answer (B)

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 20: 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) $-\frac{1}{10}$

20) Answer (A)

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

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