NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 - Inverse Trigonometric Functions

Question:2 Prove the following: $3\cos^{-1} x = \cos^{-1}(4x^3 - 3x), \;\;x\in\left[\frac{1}{2},1 \right ]$

Answer:

Given to prove $3\cos^{-1} x = \cos^{-1}(4x^3 - 3x), \;\;x\in\left[\frac{1}{2},1 \right ]$ .

Take $\cos^{-1}x = \theta$ or $\cos \theta = x$ ;

Then we have;

R.H.S.

$\cos^{-1}(4x^3 - 3x)$

= $\cos^{-1}(4\cos^3 \theta - 3\cos\theta)$ $\left [ \because 4\cos^3 \theta - 3\cos\theta = \cos3 \theta \right ]$

= $\cos^{-1}(\cos3\theta)$

= $3\theta$

= $3\cos^{-1}x$ = L.H.S

Hence Proved.

Question:3 Write the following functions in the simplest form: $\tan^{-1}\frac{\sqrt{1 + x^2}- 1}{x},\;\;x\neq 0$

Answer:

We have $\tan^{-1}\frac{\sqrt{1 + x^2}- 1}{x}$

Take

$\therefore$ $\tan^{-1} \frac {\sqrt{1+x^2} - 1}{x} = \tan^{-1}\frac{\sqrt{1+\tan^2 \Theta - 1}}{\tan \Theta}$

$=\tan^{-1}(\frac{sec \Theta-1}{tan \Theta}) = \tan^{-1}\left ( \frac{1-cos \Theta}{sin \Theta} \right )$

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

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

$=\frac{1}{2}\tan^{-1}x$ is the simplified form.

Question:4 Write the following functions in the simplest form: $\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1 + \cos x}} \right ),\;\; 0< x < \pi$

Answer:

Given that $\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1 + \cos x}} \right ),\;\; 0< x < \pi$

We have in inside the root the term : $\frac{1-\cos x}{1 + \cos x}$

Put $1-\cos x = 2\sin^2\frac{x}{2}$ and $1+\cos x = 2\cos^2\frac{x}{2}$ ,

Then we have,

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

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

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

Hence the simplest form is $\frac{x}{2}$

Question: 5 Write the following functions in the simplest form: $\tan^{-1}\left(\frac{\cos x -\sin x }{\cos x + \sin x} \right ),\;\; \frac{-\pi}{4} < x < \frac{3\pi}{4}$

Answer:

Given $\tan^{-1}\left(\frac{\cos x -\sin x }{\cos x + \sin x} \right )$ where $x\:\epsilon\:( \frac{-\pi}{4} < x < \frac{3\pi}{4})$

So,

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

Taking $\cos x$ common from numerator and denominator.

We get:

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

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

= $\tan^{-1}(1) - \tan^{-1}(\tan x)$ as, $\left [ \because \tan^{-1}x - \tan^{-1}y = \frac{x - y}{1 + xy} \right ]$

= $\frac{\pi}{4} - x$ is the simplest form.

Question:6 Write the following functions in the simplest form: $\tan^{-1} \frac{x}{\sqrt{a^2 - x^2}}, \;\; |x| < a$

Answer:

Given that $\tan^{-1} \frac{x}{\sqrt{a^2 - x^2}}, \;\; |x| < a$

Take $x = a\sin \theta$ or

$\theta = \sin^{-1}\left ( \frac{x}{a} \right )$ and putting it in the equation above;

$\tan^{-1} \frac{a\sin \theta}{\sqrt{a^2 - (a\sin \theta)^2}}$

$=\tan^{-1} \frac{a\sin \theta}{a\sqrt{1 - \sin^2 \theta}}$

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

$=\tan^{-1}\left ( \tan \theta \right )$

$=\theta = \sin^{-1}\left ( \frac{x}{a} \right )$ is the simplest form.

Question:7 Write the following functions in the simplest form: $\tan^{-1}\left(\frac{3a^2x -x^3}{a^3 - 3ax^2} \right ),\;\;a>0\;\;;\;\;\frac{-a}{\sqrt3} < x < \frac{a}{\sqrt3}$

Answer:

Given $\tan^{-1}\left(\frac{3a^2x -x^3}{a^3 - 3ax^2} \right )$

Here we can take $x = a\tan \theta \Rightarrow \frac{x}{a} = \tan \theta$

So, $\theta = \tan^{-1}\left ( \frac{x}{a} \right )$

$\tan^{-1}\left(\frac{3a^2x -x^3}{a^3 - 3ax^2} \right )$ will become;

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

and as $\left [ \because \left(\frac{3\tan \theta -\tan ^3 \theta}{ 1- 3\tan ^2 \theta} \right) =\tan 3\theta \right ]$ ;

$=3 \theta$

$=3 \tan^{-1}(\frac{x}{a})$

hence the simplest form is $3 \tan^{-1}(\frac{x}{a})$ .

Question:8 Find the values of each of the following: $\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2} \right ) \right ]$

Answer:

Given equation:

$\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2} \right ) \right ]$

So, solving the inner bracket first, we take the value of $\sin x^{-1} \frac{1}{2} = x.$

Then we have,

$\sin x = \frac{1}{2} = \sin \left ( \frac{\pi}{6} \right )$

Therefore, we can write $\sin^{-1} \frac{1}{2} = \frac{\pi}{6}$ .

$\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2} \right ) \right ] = \tan^{-1}\left[2\cos\left(2\times\frac{\pi}{6} \right ) \right ]$

$= \tan^{-1}\left[2\cos\left(\frac{\pi}{3} \right ) \right ] = \tan^{-1}\left[2\times\left(\frac{1}{2} \right ) \right ] = \tan^{-1}1 = \frac{\pi}{4}$ .

Question:9 Find the values of each of the following: $\tan \frac{1}{2}\left[\sin^{-1}\frac{2x}{1+x^2} + cos^{-1}\frac{1-y^2}{1+y^2} \right ],\;\;|x|<1,\;y>0$ and $xy<1$

Answer:

Taking the value $x = \tan \Theta$ or $\tan^{-1}x = \Theta$ and $y = \tan \Theta$ or $\tan^{-1} y = \Theta$ then we have,

= $\tan \frac{1}{2}\left[\sin^{-1}\frac{2\tan \Theta}{1+(\tan \Theta)^2} + cos^{-1}\frac{1-\tan^2 \Theta}{1+(\tan \Theta)^2} \right ]$ ,

= $\tan \frac{1}{2}\left[\sin^{-1}(\sin2\Theta) + cos^{-1} (\cos 2\Theta) \right ]$

$\because \left[\cos^{-1}(\frac{1-\tan^2 \Theta}{1+ \tan^2\Theta}) = \cos^{-1} (\cos2 \Theta) , \right ]$

$\because \left[\sin^{-1}(\frac{2\tan\Theta}{1+ \tan^2\Theta}) = \sin^{-1} (\sin2 \Theta) \right ]$

Then,

$=\tan \frac{1}{2}\left[2\tan^{-1}x + 2\tan^{-1}y \right ]$ $\because \left[\tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1- xy} \right]$

$=\tan \left [ \tan^{-1}\frac{x+y}{1-xy} \right ]$

$=\frac{x+y}{1-xy}$ Ans.

Question:10 Find the values of each of the expressions in Exercises 16 to 18. $\sin^{-1}\left (\sin\frac{2\pi}{3} \right )$

Answer:

Given $\sin^{-1}\left (\sin\frac{2\pi}{3} \right )$ ;

We know that $\sin^{-1}(\sin x) = x$

If the value of x belongs to $\left [ \frac{-\pi}{2},\frac{\pi}{2} \right ]$ then we get the principal values of $\sin^{-1}x$ .

Here, $\frac{2\pi}{3} \notin \left [ \frac{-\pi}{2},\frac{\pi}{2} \right ]$

We can write $\sin^{-1}\left (\sin\frac{2\pi}{3} \right )$ is as:

= $\sin^{-1}\left [ \sin\left ( \pi-\frac{2\pi}{3} \right ) \right ]$

= $\sin^{-1}\left [ \sin \frac{\pi}{3} \right ]$ where $\frac{\pi}{3} \epsilon \left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ]$

$\therefore \sin^{-1}\left (\sin\frac{2\pi}{3} \right )=\sin^{-1}\left [ \sin \frac{\pi}{3} \right ]=\frac{\pi}{3}$

Question:11 Find the values of each of the expressions in Exercises 16 to 18. $\tan^{-1}\left (\tan\frac{3\pi}{4} \right )$

Answer:

As we know $\tan^{-1}\left ( \tan x \right ) =x$

If $x \epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ).$ which is the principal value range of $\tan^{-1}x$ .

So, as in $\tan^{-1}\left (\tan\frac{3\pi}{4} \right )$ ;

$\frac{3\pi}{4}\notin \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$

Hence we can write $\tan^{-1}\left (\tan\frac{3\pi}{4} \right )$ as :

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

Where $-\frac{\pi}{4} \epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$

and $\therefore \tan^{-1}\left (\tan\frac{3\pi}{4} \right )=\tan^{-1}\left [ \tan (\frac{-\pi}{4}) \right ]=-\frac{\pi}{4}$

Question:12 Find the values of each of the expressions in Exercises 16 to 18. $\tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2} \right )$

Answer:

Given that $\tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2} \right )$

we can take $\sin^{-1}\frac{3}{5} = x$ ,

then $\sin x = \frac{3}{5}$

or $\cos x = \sqrt{1-\sin^{2}x}= \frac{4}{5}$

$\Rightarrow \tan x = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$

$\Rightarrow \tan^{-1}\frac{3}{4}= x$

We have similarly;

$\cot^{-1} \frac{3}{2} = \tan^{-1} \frac{2}{3}$

Therefore we can write $\tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2} \right )$

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

$=\tan\left[\tan^{-1}\left ( \frac{\frac{3}{4}+\frac{2}{3}}{1- \frac{3}{4}.\frac{2}{3}} \right ) \right ]$ from As, $\left[ \tan^{-1}x + \tan^{-1}y = \tan^{-1} \left( \frac{x+y}{1 - xy} \right) \right]$

$=\tan \left (\tan^{-1} \frac{9+8}{12-6} \right ) = \tan \left (\tan^{-1} \frac{17}{6} \right )= \frac{17}{6}$

Question:13 $\cos^{-1}\left(\cos\frac{7\pi}{6} \right )$ is equal to

(A) $\frac{7\pi}{6}$

(B) $\frac{5\pi}{6}$

(C) $\frac{\pi}{3}$

(D) $\frac{\pi}{6}$

Answer:

As we know that $\cos^{-1} (cos x ) = x$ if $x\epsilon [0,\pi]$ and is principal value range of $\cos^{-1}x$ .

In this case $\cos^{-1}\left(\cos\frac{7\pi}{6} \right )$ ,

$\frac{7\pi}{6} \notin [0,\pi]$

hence we have then,

$\cos^{-1}\left(\cos\frac{7\pi}{6} \right ) =$ $\cos^{-1} \left ( \cos \frac{-7\pi}{6} \right ) = \cos^{-1}\left [ \cos\left ( 2\pi - \frac{7\pi}{6} \right ) \right ]$

$\left [ \because \cos (2\pi + x) = \cos x \right ]$

Therefore we have $\cos^{-1}\left( \cos \frac{7\pi}{6} \right) = \cos^{-1}\left( \cos \frac{5\pi}{6} \right) = \frac{5\pi}{6}$

Hence the correct answer is $\frac{5\pi}{6}$ (B).

Question:14 $\sin\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right )$ is equal to

(A) $\frac{1}{2}$

(B) $\frac{1}{3}$

(C) $\frac{1}{4}$

(D) $1$

Answer:

Solving the inner bracket of $\sin\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right )$ ;

$\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right )$ or

Take $\sin^{-1}\left(-\frac{1}{2} \right ) = x$ then,

$\sin x =-\frac{1}{2}$ and we know the range of principal value of $\sin^{-1}x\ is\ \left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ].$

Therefore we have $\sin^{-1}\left ( -\frac{1}{2} \right ) = -\frac{\pi}{6}$ .

Hence, $\sin\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right ) = \sin \left ( \frac{\pi}{3}+ \frac{\pi}{6} \right )= \sin \left ( \frac{3\pi}{6} \right ) = \sin\left ( \frac{\pi}{2} \right ) = 1$

Hence the correct answer is D.

Question:15 $\tan^{-1}\sqrt3 - \cot^{-1}(-\sqrt3)$ is equal to

(A) $\pi$

(B) $-\frac{\pi}{2}$

(C) 0

(D) $2\sqrt3$

Answer:

We have $\tan^{-1}\sqrt3 - \cot^{-1}(-\sqrt3)$ ;

finding the value of $\cot^{-1}(-\sqrt3)$ :

Assume $\cot^{-1}(-\sqrt3) =y$ then,

$\cot y = -\sqrt 3$ and the range of the principal value of $\cot^{-1}$ is $(0,\pi)$ .

Hence, principal value is $\frac{5\pi}{6}$

Therefore $\cot^{-1} (-\sqrt3) = \frac {5\pi}{6}$

and $\tan^{-1} \sqrt3 = \frac{\pi}{3}$

so, we have now,

$\tan^{-1}\sqrt3 - \cot^{-1}(-\sqrt3)=\frac{\pi}{3} - \frac{5\pi}{6}$

$= \frac{2\pi - 5\pi}{6} = \frac{-3\pi}{6}$

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

Hence the answer is option (B).