XAT Trignometry Questions PDF [Most Important]
Download Trignometry Questions for XAT PDF – XAT Trignometry questions pdf by Cracku. Top 12 very Important Trignometry Questions for XAT based on asked questions in previous exam papers.
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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 $3 \sec \theta + 4 \cos \theta – 4\sqrt{3} = 0$ where $\theta$ is an acute angle then the value of $\theta$ is:
a) $20^\circ$
b) $30^\circ$
c) $60^\circ$
d) $45^\circ$
3) 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 4: 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) $-\frac{3}{2}$
4) 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
Question 5: 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) $\sqrt[4]{2}$
5) 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 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
