NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions

Q1. (i) Find the principal values of the following: ${\sin }^{-1}\left(\frac{-1}{2}\right)$

Answer:

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

$⟹\sin x=\frac{-1}{2}=-\sin (\frac{\pi }{6})=\sin (-\frac{\pi }{6})$
We know, principle value range of $si{n}^{-1}$ is $[-\frac{\pi }{2},\frac{\pi }{2}]$

$\therefore$ The principal value of ${\sin }^{-1}\left(\frac{-1}{2}\right)$ is $-\frac{\pi }{6},$

Q1. (ii) Find the principal values of the following: ${\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Answer:

So, let us assume that ${\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)=x$ then,

Taking inverse both sides we get;

$cosx=(\frac{\sqrt{3}}{2})$ , or $cos(\frac{\pi }{6})=(\frac{\sqrt{3}}{2})$

and as we know that the principal values of $co{s}^{-1}$ is from [0, $\pi$ ],

Hence $cosx=(\frac{\sqrt{3}}{2})$ when x = $\frac{\pi }{6}$ .

Therefore, the principal value for ${\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)$ is $\frac{\pi }{6}$ .

Q1. (iii) Find the principal values of the following:${\text{cosec}}^{-1}(2)$

Answer:

Let us assume that ${\text{cosec}}^{-1}(2)=x$ , then we have;

$Cosecx=2$ , or

$Cosec(\frac{\pi }{6})=2$ .

And we know the range of principal values is $[\frac{-\pi }{2},\frac{\pi }{2}]-\left\{0\right\}.$

Therefore the principal value of ${\text{cosec}}^{-1}(2)$ is $\frac{\pi }{6}$ .

Q1. (iv)Find the principal values of the following:${\tan }^{-1}(-\sqrt{3})$

Answer:

Let us assume that ${\tan }^{-1}(-\sqrt{3})=x$ , then we have;

$\tan x=(-\sqrt{3})$ or

$-\tan (\frac{\pi }{3})=\tan \left(\frac{-\pi }{3}\right).$

and as we know that the principal value of ${\tan }^{-1}$ is $(\frac{-\pi }{2},\frac{\pi }{2})$ .

Hence the only principal value of ${\tan }^{-1}(-\sqrt{3})$ when $x=\frac{-\pi }{3}$ .

Q1. (v)Find the principal values of the following: ${\cos }^{-1}(-\frac{1}{2})$

Answer:

Let us assume that ${\cos }^{-1}(-\frac{1}{2})=y$ then,

Easily we have; $\cos y=\left(\frac{-1}{2}\right)$ or we can write it as:

$-\cos \left(\frac{\pi }{3}\right)=\cos (\pi -\frac{\pi }{3})=\cos \left(\frac{2\pi }{3}\right).$

as we know that the range of the principal values of ${\cos }^{-1}$ is $[0,\pi ]$

Hence $\frac{2\pi }{3}$ lies in the range it is a principal solution.

Q1. (vi)Find the principal values of the following ${\tan }^{-1}(-1)$

Answer:

Given ${\tan }^{-1}(-1)$we can assume it to be equal to 'z';

${\tan }^{-1}(-1)=z$ ,

$\tan z=-1$

or

$-\tan (\frac{\pi }{4})=\tan (\frac{-\pi }{4})=-1$

And as we know the range of principal values of ${\tan }^{-1}$ from $(\frac{-\pi }{2},\frac{\pi }{2})$ .

As only one value z = $-\frac{\pi }{4}$ lies hence we have only one principal value that is $-\frac{\pi }{4}$ .

Q1. (vii)Find the principal values of the following: ${\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Answer:

Let us assume that ${\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)=z$ then,

we can also write it as; $\sec z=\left(\frac{2}{\sqrt{3}}\right)$ .

Or $\sec (\frac{\pi }{6})=\left(\frac{2}{\sqrt{3}}\right)$ and the principal values lies between $[0,\pi ]-\left\{\frac{\pi }{2}\right\}$ .

Hence we get only one principal value of ${\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)$ i.e., $\frac{\pi }{6}$ .

Q1. (viii) Find the principal values of the following: ${\cot }^{-1}(\sqrt{3})$

Answer:

Let us assume that ${\cot }^{-1}(\sqrt{3})=x$ , then we can write in other way,

$\cot x=(\sqrt{3})$ or

$\cot (\frac{\pi }{6})=(\sqrt{3})$ .

Hence when $x=\frac{\pi }{6}$ we have $\cot (\frac{\pi }{6})=(\sqrt{3})$ .

and the range of principal values of ${\cot }^{-1}$ lies in $(0,\pi )$ .

Then the principal value of ${\cot }^{-1}(\sqrt{3})$ is $\frac{\pi }{6}$

Q1. (ix) Find the principal values of the following: ${\cos }^{-1}(-\frac{1}{\sqrt{2}})$

Answer:

Let us assume ${\cos }^{-1}(-\frac{1}{\sqrt{2}})=x$ ;

Then we have $\cos x=\left(\frac{-1}{\sqrt{2}}\right)$

or

$-\cos (\frac{\pi }{4})=\left(\frac{-1}{\sqrt{2}}\right)$ ,

$\cos (\pi -\frac{\pi }{4})=\cos (\frac{3\pi }{4})$ .

And we know the range of principal values of ${\cos }^{-1}$ is $[0,\pi ]$.

So, the only principal value which satisfies ${\cos }^{-1}(-\frac{1}{\sqrt{2}})=x$ is $\frac{3\pi }{4}$ .

Q1. (x) Find the principal values of the following: ${\text{cosec}}^{-1}(-\sqrt{2})$

Answer:

Let us assume the value of ${\text{cosec}}^{-1}(-\sqrt{2})=y$ , then

we have $cosecy=(-\sqrt{2})$or

$-cosec(\frac{\pi }{4})=(-\sqrt{2})=cosec(\frac{-\pi }{4})$ .

and the range of the principal values of ${\text{cosec}}^{-1}$ lies between $[\frac{-\pi }{2},\frac{\pi }{2}]-\left\{0\right\}$ .

hence the principal value of ${\text{cosec}}^{-1}(-\sqrt{2})$ is $\frac{-\pi }{4}$ .

Question:11 Find the values of the following: ${\tan }^{-1}(1)+{\cos }^{-1}(-\frac{1}{2})+{\sin }^{-1}(-\frac{1}{2})$

Answer:

To find the values first we declare each term to some constant ;

$ta{n}^{-1}(1)=x$ , So we have $\tan x=1$ ;

or $\tan (\frac{\pi }{4})=1$

Therefore, $x=\frac{\pi }{4}$

$co{s}^{-1}(\frac{-1}{2})=y$

So, we have

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

Therefore $y=\frac{2\pi }{3}$ ,

${\sin }^{-1}(\frac{-1}{2})=z$ ,

So we have;

$\sin z=\frac{-1}{2}$ or $-\sin (\frac{\pi }{6})=\sin (\frac{-\pi }{6})=\frac{-1}{2}$

Therefore $z=-\frac{\pi }{6}$

Hence we can calculate the sum:

$=\frac{\pi }{4}+\frac{2\pi }{3}-\frac{\pi }{6}$

$=\frac{3\pi +8\pi -2\pi }{12}=\frac{9\pi }{12}=\frac{3\pi }{4}$ .

Question:12 Find the values of the following: ${\cos }^{-1}\left(\frac{1}{2}\right)+2{\sin }^{-1}\left(\frac{1}{2}\right)$

Answer:

Here we have ${\cos }^{-1}\left(\frac{1}{2}\right)+2{\sin }^{-1}\left(\frac{1}{2}\right)$

let us assume that the value of

${\cos }^{-1}\left(\frac{1}{2}\right)=x,and{\sin }^{-1}\left(\frac{1}{2}\right)=y$ ;

then we have to find out the value of+2y.

Calculation of x :

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

$\Rightarrow \cos x=\frac{1}{2}$

$\Rightarrow \cos \frac{\pi }{3}=\frac{1}{2}$ ,

Hence $x=\frac{\pi }{3}$ .

Calculation of y :

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

$\Rightarrow \sin y=\frac{1}{2}$

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

Hence $y=\frac{\pi }{6}$ .

The required sum will be = $\frac{\pi }{3}+2(\frac{\pi }{6})=\frac{2\pi }{3}$ .

Question:13 If ${\sin }^{-1}x=y$ then

(A) $0\le y\le \pi$

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

(C) $0<y<\pi$

(D) $-\frac{\pi }{2}<y<\frac{\pi }{2}$

Answer:

Given if ${\sin }^{-1}x=y$ then,

As we know that the ${\sin }^{-1}$ can take values between $[\frac{-\pi }{2},\frac{\pi }{2}].$

Therefore, $-\frac{\pi }{2}\le y\le \frac{\pi }{2}$ .

Hence answer choice (B) is correct.

Question:14 ${\tan }^{-1}(\sqrt{3})-{\sec }^{-1}(-2)$ is equal to

(A) $\pi$

(B) $-\frac{\pi }{3}$

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

(D) $\frac{2\pi }{3}$

Answer:

Let us assume the values of ${\tan }^{-1}(\sqrt{3})$ be 'x' and ${\sec }^{-1}(-2)$ be 'y'.

Then we have;

${\tan }^{-1}(\sqrt{3})=x$ or $\tan x=\sqrt{3}$ or $\tan \frac{\pi }{3}=\sqrt{3}$ or

$x=\frac{\pi }{3}$ .

and ${\sec }^{-1}(-2)=y$or$\sec y=-2$

or $-\sec (\frac{\pi }{3})=\sec (\pi -\frac{\pi }{3})=\sec \frac{2\pi }{3}$

$y=\frac{2\pi }{3}$

also, the ranges of the principal values of ${\tan }^{-1}$ and ${\sec }^{-1}$ are $(\frac{-\pi }{2},\frac{\pi }{2})$ . and

$[0,\pi ]-\left\{\frac{\pi }{2}\right\}$ respectively.

$\therefore$ we have then;

${\tan }^{-1}(\sqrt{3})-{\sec }^{-1}(-2)$

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

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

= $\frac{\pi }{4}-x$ 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:7Write 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:11Find 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