NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions

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

Answer:

Given to prove: $3{\sin }^{-1}x={\sin }^{-1}(3x-4x^3)$

where, $xϵ[-\frac{1}{2},\frac{1}{2}]$.

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

Take R.H.S value

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

= ${\sin }^{-1}(3\sin \theta -4{\sin }^3\theta )$

= ${\sin }^{-1}(\sin 3\theta )$

= $3\theta$

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

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

Answer:

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

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

Then we have;

R.H.S.

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

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

= ${\cos }^{-1}(\cos 3\theta )$

= $3\theta$

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

Hence, Proved.

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

Answer:

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

Take

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

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

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

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

$=\frac{1}{2}{\tan }^{-1}x$ 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 $xϵ(\frac{-\pi }{4}<x<\frac{3\pi }{4})$

So,

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

Taking $\cos x$ common from the numerator and the 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, $[\because {\tan }^{-1}x-{\tan }^{-1}y=\frac{x-y}{1+xy}]$

= $\frac{\pi }{4}-x$ 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;

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

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

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

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

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

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

Answer:

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

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

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

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

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

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

$=3\theta$

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

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

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

Answer:

Given equation:

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

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 ${\sin }^{-1}\frac{1}{2}=\frac{\pi }{6}$ .

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

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

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

Answer:

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

= $\tan \frac{1}{2}[{\sin }^{-1}\frac{2\tan \mathrm{\Theta }}{1+(\tan \mathrm{\Theta }{)}^2}+co{s}^{-1}\frac{1-{\tan }^2\mathrm{\Theta }}{1+(\tan \mathrm{\Theta }{)}^2}]$ ,

= $\tan \frac{1}{2}[{\sin }^{-1}(\sin 2\mathrm{\Theta })+co{s}^{-1}(\cos 2\mathrm{\Theta })]$

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

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

Then,

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

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

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

Question 10: If ${\tan }^{-1}\frac{x-1}{x-2}+{\tan }^{-1}\frac{x+1}{x+2}=\frac{\pi }{4}$ , then find the value of $x$ .

Answer:

Using the identity ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy}$ ,

We can find the value of x.

So, ${\tan }^{-1}\frac{x-1}{x-2}+{\tan }^{-1}\frac{x+1}{x+2}=\frac{\pi }{4}$

on applying,

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

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

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

= $2x^2=1$ or $x=\pm \frac{1}{\sqrt{2}}$ ,

Hence, the possible values of x are $\pm \frac{1}{\sqrt{2}}$ .

Question 11: Find the values of each of the expressions${\sin }^{-1}(\sin \frac{2\pi }{3})$

Answer:

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

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

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

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

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

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

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

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

Question 12: Find the values of each of the expressions ${\tan }^{-1}(\tan \frac{3\pi }{4})$

Answer:

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

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

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

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

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

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

Where $-\frac{\pi }{4}ϵ(-\frac{\pi }{2},\frac{\pi }{2})$

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

Question 13: Find the values of each of the expressions $\tan ({\sin }^{-1}\frac{3}{5}+{\cot }^{-1}\frac{3}{2})$

Answer:

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

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 similarities

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

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

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

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

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

Question 14: ${\cos }^{-1}(\cos \frac{7\pi }{6})$ 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 ${\cos }^{-1}(cosx)=x$ if $xϵ[0,\pi ]$ and is principal value range of ${\cos }^{-1}x$ .

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

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

hence we have then,

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

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

$\therefore wehave{\cos }^{-1}(\cos \frac{7\pi }{6})={\cos }^{-1}(\cos \frac{5\pi }{6})=\frac{5\pi }{6}$

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

Question 15: $\sin (\frac{\pi }{3}-{\sin }^{-1}(-\frac{1}{2}))$ is equal to

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

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

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

(D) $1$

Answer:

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

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

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

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

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

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

Hence, the correct answer is D.

Question 15: ${\tan }^{-1}\sqrt{3}-{\cot }^{-1}(-\sqrt{3})$ is equal to

(A) $\pi$

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

(C) 0

(D) $2\sqrt{3}$

Answer:

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

finding the value of ${\cot }^{-1}(-\sqrt{3})$ :

Assume ${\cot }^{-1}(-\sqrt{3})=y$ 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 ${\cot }^{-1}(-\sqrt{3})=\frac{5\pi }{6}$

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

So, we have now,

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

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

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

Hence, the answer is option (B).