SNAP Trignometry Questions PDF [Important]

Trignometry is an important topic in the Quant section of the SNAP Exam. Quant is a scoring section in SNAP, so it is advised to practice as much as questions from quant. This article provides some of the most important Trignometry Questions for SNAP. One can also download this Free Trignometry Questions for SNAP PDF with detailed answers by Cracku. These questions will help you practice and solve the Trignometry questions in the SNAP exam. Utilize this PDF practice set, which is one of the best sources for practicing.

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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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