MCQ | Class 12 | Chapter 2
INVERSE TRIGONOMETRIC FUNCTIONS

MCQ | CHAPTER 2 | CLASS 12

Q 1) The value of $tan^{-1}\left [ sin\left ( \frac{-\pi }{2{\color{Orchid} }} \right ) \right ]$ is =

a) -1 b) 1 $c)\: \: \pi ^{2}$ $d)\: \: \frac{-\pi }{4}$

Ans: d
Q 2) The value of $cos^{-1}\left ( \frac{1}{2} \right )+2sin^{-1}\left ( \frac{1}{2} \right )$ is equal to

$a)\: \: \frac{\pi }{4}$ $b)\: \: \frac{\pi }{6}$ $c)\: \: \frac{\pi }{3}$ $d)\: \: \frac{2\pi }{3}$

Ans: d

Q 3) The value of sin^-1{cos (4095^o) is equal to

$a)\: \: \frac{-\pi }{3}$ $b)\: \: \frac{\pi }{6}$ $c)\: \: \frac{-\pi }{4}$ $d)\: \: \frac{\pi }{4}$

Ans: c

Q 4) The value of $cos(tan^{-1}x)\: =$

$a)\: \: \sqrt{1+x^{2}}$

$b)\: \: \frac{1}{\sqrt{1+x^{2}}}$

$c)\: \: 1+x^{2}$

d) None of these

Ans: b

Q 5) The value of $tan\: (sex^{-1}\sqrt{1+x^{2}})\: \: is$

$a)\: \: \frac{1}{x}$

b) x
$c)\: \: \: \frac{1}{\sqrt{1+x^{2}}}$

$d)\: \: \: \frac{x}{\sqrt{1+x^{2}}}$
Ans: b

Q 6) The value of $sec^{-1}[sec(-30^{o})]\: \: is=$
a) - 60 b) -30 c) 30 d) 150
Ans: c
Q 7) The value of $tan^{-1}\left [ \frac{cosx}{1+sinx} \right ]=$
$a)\: \: \: \frac{\pi }{4}-\frac{x}{2}$

$b)\: \: \: \frac{\pi }{4}+\frac{x}{2}$

$c)\: \: \: \frac{x}{2}$

$d)\: \: \: \frac{\pi }{4}-x$
Ans: a
Q 8) The value of $tan^{-1}\left ( \frac{1}{\sqrt{x^{2}-1}} \right )\: \: is=$

$a)\: \: \: \frac{\pi }{2}+cosec^{-1}x$

$b)\: \: \: \frac{\pi }{2}+sec^{-1}x$

$c)\: \: \: cosec^{-1}x$

$d)\: \: \: sec^{-1}x$
Ans: c
Q 9) The value of $cos^{-1}\left ( cos\frac{7\pi }{5} \right )$ is =

$a)\: \: \: \frac{3\pi }{5}$ $b)\: \: \: \frac{2\pi }{5}$ $c)\: \: \: \frac{-7\pi }{5}$ $d)\: \: \: \frac{7\pi }{5}$

Ans: a
Q 10) The principal value of $sin^{-1}\left ( \frac{-1}{2} \right )\: \: is=$

$a)\: \: \: \frac{\pi }{3}$ $b)\: \: \: \frac{\pi }{6}$ $c)\: \: \: \frac{-\pi }{3}$ $d)\: \: \: \frac{-\pi }{6}$

Ans: d

Q 11) Sec²(tan^-12) + cosec²(cot^-13) =

a) 5 b) 13 c) 15 d) 16
Ans: c

Q 12) $sin^{-1}\left [x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^{2}} \right ] =$

$a)\: \: sin^{-1}x+sin^{-1}\sqrt{x}$

$b)\: \: sin^{-1}x-sin^{-1}\sqrt{x}$

$c)\: \: sin^{-1}\sqrt{x}-sin^{-1}x$

d) None of these

Ans: b

Q 13) $sin\left [ \frac{1}{2}cot^{-1}\left ( \frac{-3}{4} \right ) \right ]\: \: is=$
$a)\: \: \frac{1}{\sqrt{5}}$

$b)\: \: \frac{-2}{\sqrt{5}}$

$c)\: \: \frac{2}{\sqrt{5}}$

$d)\: \: \frac{-1}{\sqrt{5}}$

Ans: c

Q 14) If $tan^{-1}\left ( \frac{1-x}{1+x} \right )=\frac{1}{2}tan^{-1}x$ , then x =

a) 1

$b)\:\: \sqrt{3}$

$c)\:\: \frac{1}{\sqrt{3}}$

d) None of these

Ans: c

Q 15) The value of $cos^{-1}\left ( cos\frac{7\pi }{6} \right )$

$a)\:\:\frac{7\pi }{6}$

$b)\:\:\frac{5\pi }{6}$

$c)\:\:\frac{\pi }{6}$

d) None of these

Ans: b

Q 16) If $cos^{-1}\left ( \frac{5}{13} \right )-sin^{-1}\left ( \frac{12}{13} \right )=cos^{-1}x$ , then x is equal to

a) 1 $b)\:\:\frac{1}{\sqrt{2}}$ c) 0 $d)\:\:\frac{\sqrt{3}}{2}$

Ans: a

Q 17) Principal value of $sin^{-1}\left ( \frac{-1}{2} \right )$ is

$a)\: \: \frac{\pi }{3}$ $b)\: \: \frac{-\pi }{3}$ $c)\: \: \frac{5\pi }{6}$ $d)\: \: \frac{-\pi }{6}$

Ans: d

Explanation:

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

$\Rightarrow \: \: sin\theta =\frac{-1}{2}=sin\left ( \frac{-\pi }{6} \right )$

$\Rightarrow \: \: \theta =\frac{-\pi }{6}$

Q 18) If $sin^{-1}x+sin^{-1}y=\frac{2\pi }{3}$ , then find the value of $cos^{-1}x+cos^{-1}y$
$a)\: \: \frac{2\pi }{3}$

$b)\: \: \frac{-2\pi }{3}$

$c)\: \: \frac{\pi }{3}$

$d)\: \: \pi$

Ans: c

Explanation

$sin^{-1}x+sin^{-1}y=\frac{2\pi }{3}$

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

$\pi -cos^{-1}x-cos^{-1}y=\frac{2\pi }{3}$

$cos^{-1}x+cos^{-1}y=\pi -\frac{2\pi }{3}$

$cos^{-1}x+cos^{-1}y=\frac{\pi }{3}$

Q 19) $tan^{-1}\left [ sin\left ( \frac{-\pi }{2} \right ) \right ]$ is equal to

a) -1 b) 1 $c)\: \: \frac{\pi }{2}$ $d)\: \: \frac{-\pi }{4}$
Ans: d

Q 20) $sin\left [ cot^{-1}\left\{ cos\left ( tan^{-1}x \right )\right\} \right ] =$

$a)\: \: \sqrt{\frac{1+x^{2}}{2+x^{2}}}$

$b)\: \: \sqrt{\frac{1-x^{2}}{2+x^{2}}}$

$c)\: \: \sqrt{\frac{1+x^{2}}{2-x^{2}}}$

$d)\: \: \sqrt{\frac{2+x^{2}}{1+x^{2}}}$

Ans: a

Q21) The value of $sec\left\{tan^{-1}\left ( \frac{-y}{3} \right ) \right\}$ is equal to

a) $\frac{\sqrt{9+y^{2}}}{9}$

b) $\frac{\sqrt{9+y^{2}}}{3}$

c) $\frac{3}{\sqrt{9+y^{2}}}$

d) $\frac{9}{\sqrt{9+y^{2}}}$

Ans b

Explanation

$sec\left ( tan^{-1}\frac{y}{3} \right )=\sqrt{sec^{2}\left ( tan^{-1}\frac{y}{3} \right )}$

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

$=\sqrt{1+\frac{y^{2}}{9}}=\frac{\sqrt{9+y^{2}}}{3}$

Q22) If $sin^{-1}x+sin^{-1}y+sin^{-1}z = \frac{3\pi }{2}$ then the value of x + y² + z³is

a) 1 b) 3 c) 2 d) 5

Ans: b

Explanation

$sin^{-1}x+sin^{-1}y+sin^{-1}z = \frac{3\pi }{2}$

$\Rightarrow sin^{-1}x=\frac{\pi}{2}$ $\Rightarrow x=1$

$\Rightarrow sin^{-1}y=\frac{\pi}{2}$ $\Rightarrow y=1$

$\Rightarrow sin^{-1}z=\frac{\pi}{2}$ $\Rightarrow z=1$

$x+y^{2}+z^{3}=1+1+1=3$

Q23 Principal value of the expression $cos^{-1}\left [ cos\left ( -680^{o} \right ) \right ]$ is

a) $\frac{2\pi }{9}$

b) $-\frac{2\pi }{9}$

c) $\frac{34\pi }{9}$

d) $\frac{\pi }{9}$

Ans : a

Q24) The value of tan²(sec^-12) + cot²(cosec^-13) is

a) 5 b) 11 c) 13 d) 15

Ans: b

Q25) The value of $tan^{-1}\left ( \frac{1}{2} \right )+tan^{-1}\left ( \frac{1}{3} \right )+tan^{-1}\left ( \frac{7}{8} \right )$ is

a) $tan^{-1}\left ( \frac{7}{8} \right )$

b) $cot^{-1}(15)$

c) $tan^{-1}(15)$

d) $tan^{-1}\left ( \frac{25}{24} \right )$

Ans : c

Q26) Find the value of $tan^{-1}\left ( \sqrt{3} \right )-cot^{-1}\left ( -\sqrt{3} \right )$

a) π

b) $-\frac{\pi }{2}$

c) - π

d) $\frac{\pi }{2}$

Answer: b

Explanation

$tan^{-1}\sqrt{3}-cot^{-1}(-\sqrt{3})$

$=tan^{-1}\sqrt{3}-\left [\pi - cot^{-1}(\sqrt{3}) \right ]$

$=\left [ tan^{-1}\sqrt{3}+cot^{-1}(\sqrt{3}) \right ]-\pi$

$=\frac{\pi }{2}-\pi =-\frac{\pi }{2}$

Q27) If sec^-1x + sec^-1y = $\frac{\pi }{2}$ , then the value of cosec^-1x + cosec^-1y is

a) π b) π/2 c) 3π/2 d) -π

Answer b

Explanation

sec^-1x + sec^-1y = $\frac{\pi }{2}$

⇒ $\frac{\pi }{2}$ - cosec^-1x + $\frac{\pi }{2}$ - cosec^-1y = $\frac{\pi }{2}$

⇒ $\frac{\pi }{2}$ - cosec^-1x - cosec^-1y = 0

⇒ cosec^-1x + cosec^-1y = $\frac{\pi }{2}$

Q28) $cot\left ( \frac{\pi }{4}-2cot^{-1}3 \right )=$

a) 5 b) 9 c) 12 d) 7

Answer 7

Explanation

$2cot^{-1}(3)=2tan^{-1}\left ( \frac{1}{3} \right )$

= $tan^{-1}\left ( \frac{\frac{2}{3}}{1-\frac{1}{9}} \right )=tan^{-1}\left ( \frac{2/3}{8/9} \right )$

$=tan^{-1}\left ( \frac{2\times 9}{3\times 8} \right )=tan^{-1}\left ( \frac{3}{4} \right )$

Now by using the formula

$cot\left(\frac{\pi}{4}-\theta\right)=\frac{1+tan\theta}{1-tan\theta}$

Given equations becomes

$cot\left ( \frac{\pi }{4}-2cot^{-1}3 \right )=cot\left ( \frac{\pi }{4}-tan^{-1}\frac{3}{4} \right )$

$cot\left ( \frac{\pi }{4}-tan^{-1}\frac{3}{4} \right )=\frac{1+tan\left(tan^{-1}\frac{3}{4}\right)}{1-tan\left(tan^{-1}\frac{3}{4}\right)}$

$\frac{1+tan\left(tan^{-1}\frac{3}{4}\right)}{1-tan\left(tan^{-1}\frac{3}{4}\right)}=\frac{1+\frac{3}{4}}{1-\frac{3}{4}}=7$

Q29) If 3sin^-1x + cos^-1x = π, then x is equal to

a) 0

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

c) -1

d) 1/2

Answer b

Explanation

3sin^-1x + cos^-1x = π

2sin^-1x + (sin^-1x + cos^-1x) = π

2sin^-1x + π / 2 = π

2sin^-1x = π - π / 2 = π / 2

sin^-1x = π / 4

x = sin(π / 4) = $\frac{1}{\sqrt{2}}$

Q30) If $sin^{-1}\left ( \frac{2\alpha }{1+\alpha ^{2}} \right )+cos^{-1}\left ( \frac{1-\beta ^{2}}{1+\beta ^{2}} \right )=tan^{-1}\left ( \frac{2x}{1-x^{2}} \right )$ , then the value of x is

a) $x=\frac{\alpha +\beta }{1-\alpha \beta }$

b) $x=\frac{\alpha -\beta }{1-\alpha \beta }$

c) $x=\frac{\alpha -\beta }{1+\alpha \beta }$

d) None of these

Ans: a

Explanation

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

$cos^{-1}\left ( \frac{1-\beta ^{2}}{1+\beta ^{2}} \right )=2tan^{-1}\beta$

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

Using these values in the given equation we get

$2tan^{-1}\alpha +2tan^{-1}\beta =2tan^{-1}x$

$tan^{-1}\alpha +tan^{-1}\beta =tan^{-1}x$

$\Rightarrow tan^{-1}\left ( \frac{\alpha +\beta }{1-\alpha \beta } \right )=tan^{-1}x$

$\Rightarrow x=\frac{\alpha +\beta }{1-\alpha \beta }$

Q31) The value of : $tan\left ( \frac{1}{2}cos^{-1}\frac{\sqrt{5}}{3} \right )$ is

a) $\frac{3-\sqrt{5}}{2}$

b) $\frac{3+\sqrt{5}}{2}$

c) $\frac{5+\sqrt{3}}{2}$

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

Answer a

Q32) The value of : $sin^{-1}\left ( \frac{2\sqrt{2}}{3} \right )+sin^{-1}\left ( \frac{1}{3} \right )$

a) $\frac{\pi }{6}$

b) $\frac{\pi }{4}$

c) $\frac{\pi }{2}$

d) 0

Answer c

Q33) The value of expression $2sec^{-1}2 + sin^{-1}\left ( \frac{1}{2} \right )$ is

a) $\frac{\pi }{6}$ b) $\frac{5\pi }{6}$ c) $\frac{7\pi }{6}$ d) 1

Answer b

Q34) The value of : $sec^{2}(tan^{-1}3)+cosec^{2}(cot^{-1}2)$ is

a) 5 b) 13 c) 15 d) 25

Answer c

Q35) $sin\left [ \frac{\pi }{3}-sin^{-1}\left ( -\frac{1}{2} \right ) \right ]$ is equal to

a) 1/2 b) 1/3 c) 1/4 d) 1

Ans d

Q36) $cos\left [ 2cos^{-1}\left ( \frac{1}{5} \right ) +sin^{-1}\left ( \frac{1}{5} \right )\right ]$ is equal to

a) 1/5

b) $-\frac{2\sqrt{6}}{5}$

c) -1/5

d) $-\frac{2\sqrt{6}}{5}$

Answer: d

Q37) The value of $sin^{-1}\left [ sin\left ( \frac{-17\pi }{8} \right ) \right ]$ is

a) $-\frac{\pi }{6}$ b) $\frac{\pi }{8}$ c) $\frac{-\pi }{8}$ d) $\frac{\pi }{12}$

Answer c

Q38) Evaluate: $sin^{-1}\left\{sin\left ( \frac{5\pi }{6} \right ) \right\}+tan^{-1}\left\{ tan\left ( \frac{\pi }{6} \right )\right\}$ is

a) $\frac{\pi }{6}$ b) $\frac{2\pi }{3}$ c) $\frac{\pi }{3}$ d) $\frac{5\pi }{6}$

Answer: c

Q39) Find the value of : $sin\left [\frac{\pi }{3}-sin^{-1}\left ( -\frac{1}{2} \right ) \right ]$

Ans: 1

Q40) $sin^{-1}\left [ sin\left ( \frac{2\pi }{3} \right ) \right ]=\frac{2\pi }{3}$ , state true or false

Ans: False

Q41) What is the Principal branch of tan^-1x

Ans $\left ( -\frac{\pi }{2},\frac{\pi }{2} \right )$

Q42) If $tan^{-1}x=sin^{-1}\left ( \frac{1}{\sqrt{2}} \right )$ then find the value of x

Ans 1

Q43) $cos^{-1}\left ( cos\frac{7\pi }{6} \right )=\frac{7\pi }{6}$ state true or false

Ans : False

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Rum TanApril 30, 2024 at 10:49 AM
Inverse trigonometry deals with finding the angle from the given ratio of sides in a right triangle. It's the opposite of regular trigonometry, which finds ratios given an angle. Functions like arcsine, arccosine, and arctangent are used to calculate these angles. They provide the angle measure corresponding to a specific trigonometric ratio. Inverse trigonometric functions are essential in various fields like engineering, physics, and navigation for solving problems involving angles and sides in triangles. They offer a way to reverse the effects of trigonometric functions, aiding in problem-solving and analysis.

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MCQ | Class 12 | Chapter 2
INVERSE TRIGONOMETRIC FUNCTIONS