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Download Trignometry Questions for XAT PDF – XAT Trignometry questions PDF by Cracku. Practice XAT solved Trignometry Questions paper tests, and these are the practice question to have a firm grasp on the Trignometry topic in the XAT exam. Top 20 very Important Trignometry Questions for XAT based on asked questions in previous exam papers.  The XAT question papers contain actual questions asked with answers and solutions.

Question 1: 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) $\frac{1}{\sqrt{3}}$

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 2: 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) $\frac{2}{5}(1 + \sqrt{3})$

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 3: 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

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 4: 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) $\sec^4 \theta$

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 5: 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

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 6: 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) $\frac{1 – \sqrt{3}}{2}$

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 7: In the given figure, $\cos\theta$ is equal to:

a) $\frac{5}{13}$

b) $\frac{12}{13}$

c) $\frac{5}{12}$

d) $\frac{12}{5}$

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 8: 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

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 9: 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

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)

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

$x^2 + y^2 = 17$

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

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

Question 10: 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

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$\degree$

$(2 \sin 45 \degree + \cosec 45 \degree) = 2 \sqrt{2}$

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

$2\sqrt{2} = 2\sqrt{2}$

$2(\sin^{4}A + \cos^{4}A)$

= $2(\sin^{4}45 \degree + \cos^{4}45\degree)$

= $2((\frac{1}{\sqrt{2}})^{4} + (\frac{1}{\sqrt{2}})^{4})$

= $2(\frac{1}{4} + \frac{1}{4}) = 2(\frac{1}{2}) = 1$