NCERT Solutions for Exercise 2.1 Class 12 Maths Chapter 2 - Inverse Trigonometric Functions

Question:1 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 $-\frac{\pi }{6},$

Question:2 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 $\frac{\pi }{6}$ .

Question:3 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 $\frac{\pi }{6}$ .

Question:4 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 $x=\frac{-\pi }{3}$ .

Question:5 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.

Question:6 Find the principal values of the following : ${\tan }^{-1}(-1)$

Answer:

Given ${\tan }^{-1}(-1)$ so 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 $-\frac{\pi }{4}$ .

Question:7 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., $\frac{\pi }{6}$ .

Question:8 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 $\frac{\pi }{6}$

Question:9 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 $\frac{3\pi }{4}$ .

Question:10 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 x +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;

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

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