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

$\implies \sin x = \frac{-1}{2}= -\sin(\frac{\pi}{6}) = \sin(-\frac{\pi}{6})$
We know, principle value range of $\sin^{-1}$ 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: $\cos^{-1}\left(\frac{\sqrt3}{2} \right )$

Answer:

So, let us assume that $\cos^{-1}\left(\frac{\sqrt3}{2} \right ) = x$ then,

Taking the inverse of both sides, we get;

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

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

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

Therefore, the principal value for $\cos^{-1}\left(\frac{\sqrt3}{2} \right )$ is $\frac{\pi}{6}$ .

Question 3: Find the principal values of the following: $\textup{cosec}^{-1}(2)$

Answer:

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

$\textup{cosec}\ x = 2$ , or

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

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

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

Question 4: Find the principal values of the following: $\tan^{-1}(-\sqrt3)$

Answer:

Let us assume that $\tan^{-1}(-\sqrt3) = x$ , 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 $\left ( \frac{-\pi}{2}, \frac{\pi}{2} \right )$ .

Hence the only principal value of $\tan^{-1}(-\sqrt3)$ when $x = \frac{-\pi}{3}$ .

Question 5: Find the principal values of the following: $\cos^{-1}\left(-\frac{1}{2} \right )$

Answer:

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

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

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

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

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 $\tan^{-1}(-1)$, 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 $\left ( \frac{-\pi}{2}, \frac{\pi}{2} \right )$ .

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: $\sec^{-1}\left (\frac{2}{\sqrt3}\right)$

Answer:

Let us assume that $\sec^{-1}\left (\frac{2}{\sqrt3}\right) = z$ then,

we can also write it as; $\sec z = \left (\frac{2}{\sqrt3}\right)$ .

Or $\sec (\frac{\pi}{6}) = \left (\frac{2}{\sqrt3}\right)$ and the principal values lies between $\left [ 0, \pi \right ] - \left \{ \frac{\pi}{2} \right \}$ .

Hence we get only one principal value of $\sec^{-1}\left (\frac{2}{\sqrt3}\right)$ i.e., $\frac{\pi}{6}$ .

Question 8: Find the principal values of the following: $\cot^{-1}(\sqrt3)$

Answer:

Let us assume that $\cot^{-1}(\sqrt3) = x$ , then we can write in other way,

$\cot x = (\sqrt3)$ or

$\cot (\frac{\pi}{6}) = (\sqrt3)$ .

Hence when $x=\frac{\pi}{6}$ we have $\cot (\frac{\pi}{6}) = (\sqrt3)$ .

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

Then the principal value of $\cot^{-1}(\sqrt3)$ is $\frac{\pi}{6}$

Question 9: Find the principal values of the following: $\cos^{-1}\left(-\frac{1}{\sqrt2} \right )$

Answer:

Let us assume $\cos^{-1}\left(-\frac{1}{\sqrt2} \right ) = x$ ;

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 $\cos^{-1}\left(-\frac{1}{\sqrt2} \right ) = x$ is $\frac{3\pi}{4}$ .

Question 10: Find the principal values of the following: $\textup{cosec}^{-1}(-\sqrt2)$

Answer:

Let us assume the value of $\textup{cosec}^{-1}(-\sqrt2) = y$ , then

we have $cosec\ y = (-\sqrt 2)$
or

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

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

Hence the principal value of $\textup{cosec}^{-1}(-\sqrt2)$ is $\frac{-\pi}{4}$.

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

Answer:

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

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

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

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

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

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) $0\leq y \leq \pi$

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

(C) $0 < y < \pi$

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

Answer:

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

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

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

Hence, answer choice (B) is correct.

Question 14: $\tan^{-1}(\sqrt3)-\sec^{-1}(-2)$ 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 $\tan^{-1}(\sqrt3)$ be 'x' and $\sec^{-1}(-2)$ be 'y'.

Then we have;

$\tan^{-1}(\sqrt3) = x$ 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;

$\tan^{-1}(\sqrt3)-\sec^{-1}(-2)$

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

Question 1: Find the value of the following: $\cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right )$

Answer:

If $x \epsilon [0,\pi]$ then $\cos^{-1}(\cos x) = x$ , which is principal value of $\cos^{-1} x$ .

So, we have $\cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right )$

$where \:\frac{13\pi}{6} \notin \left [ 0, \pi \right ].$

$Hence\: we \:can\: write\: \cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right ) \:as$

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

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

$\frac{\pi}{6}\ \epsilon \left [ 0, \pi \right ]$

Therefore, we have,

$\cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right ) = \cos^{-1}\left (\cos\left(\frac{\pi}{6} \right ) \right ) = \frac{\pi}{6}$ .

Question 2: Find the value of the following: $\tan^{-1}\left(\tan\frac{7\pi}{6} \right )$

Answer:

We have given $\tan^{-1}\left(\tan\frac{7\pi}{6} \right )$ ;

so, as we know $\tan^{-1}\left(\tan x \right ) =x \:\:if\:\:x\epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$

So, here we have $\frac{7\pi}{6} \notin \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$ .

Therefore we can write $\tan^{-1}\left(\tan\frac{7\pi}{6} \right )$ as:

$=\tan^{-1}\left(\tan \left (2\pi - \frac{5\pi}{6} \right ) \right )$ $\left [ \because \tan(2\pi - x) = -\tan x \right ]$

$=\tan^{-1}\left[-\tan \left ( \frac{5\pi}{6} \right ) \right ]$

$=\tan^{-1}\left[\tan \left (\pi- \frac{5\pi}{6} \right ) \right ]$

$=\tan^{-1}\left[\tan \left (\frac{\pi}{6} \right ) \right ]\:\:where\:\:\frac{\pi}{6} \epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$

$\therefore \tan^{-1} \left ( \tan \frac{7\pi}{6} \right ) = \tan^{-1} \left ( \tan \frac{\pi}{6} \right ) = \frac{\pi}{6}$ .

Question 3: Prove that $2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}$

Answer:

To prove: $2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}$ ;

$L.H.S=2\sin^{-1}\frac{3}{5}$

Assume that $\sin^{-1}\frac{3}{5} = x$

then we have $\sin x = \frac{3}{5}$ .

or $\cos x = \sqrt{1-\left (\frac{3}{5} \right )^2} = \frac{4}{5}$

Therefore we have

$\tan x = \frac{3}{4}\:\:or\:\:x = \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{3}{4}$

Now,

We can write L.H.S as

$2\sin^{-1}\frac{3}{5} = 2\tan^{-1}\frac{3}{4}$

$=\tan^{-1} \left [\frac{2\times\frac{3}{4}}{1- \left ( \frac{3}{4} \right )^2} \right ]$ as we know $\left [2\tan^{-1} x = \tan^{-1} \frac{2x}{1-x^2} \right ]$

$=\tan^{-1} \left [\frac{\frac{3}{2}}{\left ( \frac{16-9}{16} \right )} \right ] = \tan^{-1}\left ( \frac{3}{2}\times \frac{16}{7} \right )$

$=\tan^{-1} \frac{24}{7}=R.H.S$

L.H.S = R.H.S

Question 4: Prove that $\sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5} =\tan^{-1}\frac{77}{36}$

Answer:

Taking $\sin ^{-1} \frac{8}{17} = x$

then,

$\sin x = \frac{8}{17} \Rightarrow \cos x = \sqrt{1- \left ( \frac{8}{17} \right )^2} = \sqrt {\frac{225}{289}} = \frac{15}{17}.$

Therefore, we have-

$\tan^{-1} x = \frac{8}{15} \Rightarrow x = \tan^{-1} \frac{8}{15}$

$\therefore \sin ^{-1} \frac{8}{17} = \tan ^{-1} \frac{8}{15}$.............(1).

$Now, let\:\sin ^{-1} \frac{3}{5} = y$ ,

Then,

$\sin ^{-1} \frac{3}{5} = \tan ^{-1} \frac{3}{4}$.............(2).

So, we have now,

L.H.S.

$\sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5}$

Using equations (1) and (2), we get,

$=\tan ^{-1} \frac{8}{15} + \tan^{-1} \frac{3}{4}$

$=\tan^{-1} \frac{\frac{8}{15}+ \frac{3}{4}}{1-\frac{8}{15}\times \frac{3}{4}}$
$[\because \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1 - xy}]$

$=\tan^{-1} (\frac{32+45}{60-24})$

$=\tan^{-1} (\frac{77}{36})$

= R.H.S.

Question 5: Prove that $\cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\frac{33}{65}$

Answer:

Take $\cos^{-1}\frac{4}{5} = x$ and $\cos^{-1}\frac{12}{13} = y$ and $\cos^{-1}\frac{33}{65} = z$

Then we have,

$\cos x = \frac{4}{5}$

$\sin x = \sqrt {1- \left ( \frac {4}{5} \right )^2} = \frac {3}{5}$

Then we can write it as:

$\tan x = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$ or $x= \tan^{-1} \frac{3}{4}$

$\therefore \cos ^{-1} \frac{4}{5} = \tan^{-1} \frac{3}{4}$...............(1)

Now, $\cos^{-1}\frac{12}{13} = y$

$\cos y = \frac{12}{13} \Rightarrow$ $\sin y =\frac{5}{13}$

$\therefore \tan y = \frac{5}{12} \Rightarrow y = \tan^{-1} \frac{5}{12}$

So, $\cos^{-1}\frac{12}{13} = \tan^{-1} \frac{5}{12}$...................(2)

Also, we have similarities;

$\cos^{-1}\frac{33}{65} = z$

Then,

$\cos^{-1}\frac{33}{65} = \tan^{-1} \frac{56}{33}$...........................(3)

Now, we have

L.H.S

$\cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13}$ so, using (1) and (2), we get,

$=\tan^{-1}\frac{3}{4} + \tan^{-1}\frac{5}{12}$

$=\tan^{-1}\left ( \frac{\frac{3}{4}+ \frac{5}{12}}{1-\left ( \frac{3}{4}\times \frac{5}{12} \right )} \right )$
$\because \left [ \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1- xy} \right ]$

$=\tan^{-1}\left ( \frac{36+20}{48-15} \right )$

$=\tan^{-1}\left ( \frac{56}{33} \right )$ or we can write it as;

$=\cos^{-1}\frac{33}{65}$

= R.H.S.

Hence proved.

Question 6: Prove that $\cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5} = \sin^{-1}\frac{56}{65}$

Answer:

Converting all terms in tan form;

Let $\cos^{-1}\frac{12}{13} = x$ , $\sin^{-1}\frac{3}{5} = y$ and $\sin^{-1}\frac{56}{65} = z$ .

Now, converting all the terms:

$\cos^{-1}\frac{12}{13} = x$ or $\cos x = \frac{12}{13}$

We can write it in tan form as:

$\cos x = \frac{12}{13} \Rightarrow$ $\sin x = \frac{5}{13}$ .

$\therefore \tan x = \frac{5}{12} \Rightarrow x = \tan^{-1} \frac{5}{12}$

or $\cos^{-1}\frac{12}{13} = \tan^{-1} \frac{5}{12}$ ................(1)

$\sin^{-1}\frac{3}{5} = y$ or $\sin y = \frac{3}{5}$

We can write it in tan form as:

$\sin y = \frac{3}{5} \Rightarrow$ $\cos y = \frac{4}{5}$

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

or $\sin^{-1}\frac{3}{5} = \tan^{-1} \frac{3}{4}$ ......................(2)

Similarly, for $\sin^{-1}\frac{56}{65} = z$ ;

we have $\sin^{-1}\frac{56}{65} = \tan^{-1} \frac{56}{33}$ .............(3)

Using (1) and (2), we have L.H.S

$\cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5}$

$= \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{3}{4}$

On applying $\tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1-xy}$

We have,

$=\tan^{-1} \frac{\frac{5}{12}+\frac{3}{4}}{1-(\frac{5}{12}.\frac{3}{4})}$

$=\tan^{-1} (\frac{20+36}{48-15})$

$=\tan^{-1} (\frac{56}{33})$

$=\sin^{-1} (\frac{56}{65})$ ...........[Using (3)]

=R.H.S.

Hence proved.

Question 7: Prove that $\tan^{-1}\frac{63}{16} = \sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5}$

Answer:

Taking R.H.S;

We have $\sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5}$

Converting sin and cos terms to tan forms:

Let $\sin^{-1}\frac{5}{13} = x$ and $\cos^{-1}\frac{3}{5} = y$

now, we have $\sin^{-1}\frac{5}{13} = x$ or $\sin x = \frac{5}{13}$

$\sin x = \frac{5}{13} \:or\: \cos x =\frac{12}{13}\:or\:\tan x = \frac{5}{12}$

$\tan x = \frac{5}{12} \Rightarrow x =\tan^{-1} \frac{5}{12}$

$\therefore \sin^{-1} \frac{5}{13} = \tan^{-1} \frac{5}{12}$............(1)

Now, $\cos^{-1}\frac{3}{5} = y\Rightarrow \cos y = \frac{3}{5}$

$\cos y = \frac{3}{5} \:or\: \sin y = \frac{4}{5}\:or\:\tan y = \frac{4}{3}$

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

$\therefore \cos^{-1}\frac{3}{5} = \tan^{-1} \frac{4}{5}$................(2)

Now, using (1) and (2), we get,

R.H.S.

$\sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5} = \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{4}{3}$

$=\tan^{-1}\left ( \frac{\frac{5}{12}+\frac{4}{3}}{1- \frac{5}{12}\times \frac{4}{3}} \right )$ as we know $\left [ \tan^{-1} x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1-xy} \right ]$

so,

$= \tan^{-1} \frac{63}{16}$

equal to L.H.S

Hence proved.

Question 8: Prove that $\tan^{-1} \sqrt{x} = \frac{1}{2}\cos^{-1}\frac{1-x}{1+x},\;\;x\in [0,1]$

Answer:

By observing the square root, we will first put

$x= \tan^2 \theta$ .

Then,

we have $\tan^{-1} \sqrt{\tan^2 \theta} = \frac{1}{2}\cos^{-1}\frac{1-\tan^2 \theta}{1+\tan^2 \theta}$

or, R.H.S.

$\frac{1}{2}\cos^{-1}\frac{1-\tan^2 \theta}{1+\tan^2 \theta} = \frac{1}{2}\cos^{-1}(cos2 \theta)$

$= \frac{1}{2}\times 2\theta = \theta$ .

L.H.S.$\tan^{-1} \sqrt{\tan^2 \theta} = \tan^{-1}(\tan \theta) = \theta$

hence L.H.S. = R.H.S proved.

Question 9: Prove that $\cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right ) = \frac{x}{2},\;\;x\in\left(0,\frac{\pi}{4} \right )$

Answer:

Given that $\cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right )$

By observing, we can rationalise the fraction

$\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right )$

We get,

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

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

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

$= \cot \frac{x}{2}$

Therefore, we can write it as;

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

As L.H.S. = R.H.S.

Hence proved.

Question 10: Prove that $\tan^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \right ) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x,\;\;-\frac{1}{\sqrt2}\leq x\leq 1$

[Hint: Put $x = \cos 2\theta$ ]

Answer:

By using the Hint, we will put $x = \cos 2\theta$ ;

We get,

$=\tan^{-1}\left(\frac{\sqrt{1+\cos 2\theta} - \sqrt{1-\cos2\theta}}{\sqrt{1+\cos 2\theta} + \sqrt{1-\cos 2\theta}} \right )$

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

$=\tan^{-1}\left(\frac{\sqrt2{\cos \theta} - \sqrt2{\sin\theta}}{\sqrt2{\cos \theta} + \sqrt2{\sin\theta}} \right )$

$=\tan^{-1}\left(\frac{{\cos \theta} - {\sin\theta}}{{\cos \theta} + {\sin\theta}} \right )$ dividing numerator and denominator by $\cos \theta$ ,

We get,

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

$= \tan^{-1} 1 - \tan^{-1} (\tan \theta)$ using the formula $\left [ \tan^{-1}x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy} \right ]$

$= \frac{\pi}{4} - \theta = \frac{\pi}{4}- \frac{1}{2}\cos^{-1}x$

As L.H.S = R.H.S

Hence proved

Question 11: Solve the following equations: $2\tan^{-1}(\cos x) = \tan^{-1}(2\textup{cosec}x)$

Answer:

Given equation $2\tan^{-1}(\cos x) = \tan^{-1}(2\textup{cosec}x)$ ;

Using the formula:

$\left [ 2\tan^{-1}z = \tan^{-1} \frac{2z}{1-z^2} \right ]$

We can write

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

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

So, we can equate;

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

$=\left [ \frac{2\cos x}{\sin^2 x }\right ] = \left [ \frac{2}{sinx } \right ]$

that implies that $\cos x = \sin x$ .

or $\tan x =1$ or $x = \frac{\pi}{4}$

Hence we have solution $x = \frac{\pi}{4}$ .

Question 12: Solve the following equations: $\tan^{-1} \frac{1-x}{1+x} = \frac{1}{2}\tan^{-1}x,\;(x>0)$

Answer:

Given equation is

$\tan^{-1} \frac{1-x}{1+x} = \frac{1}{2}\tan^{-1}x$ :

L.H.S can be written as;

$\tan^{-1} \frac{1-x}{1+x} = \tan^{-1}1 - \tan^{-1}x$

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

So, we have $\tan^{-1}1 - \tan^{-1}x = \frac{1}{2} \tan^{-1}x$

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

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

$\Rightarrow \tan^{-1}x = \frac{\pi}{6}$

$\Rightarrow x= \tan \frac{\pi}{6} = \frac{1}{\sqrt3}$

Hence the value of $x= \frac{1}{\sqrt3}$ .

Question 13: $\sin(\tan^{-1}x),\;|x|<1$ is equal to

(A) $\frac{x}{\sqrt{1-x^2}}$

(B) $\frac{1}{\sqrt{1-x^2}}$

(C) $\frac{1}{\sqrt{1+x^2}}$

(D) $\frac{x}{\sqrt{1+x^2}}$

Answer:

Let $\tan^{-1}x = y$ then we have;

$\tan y = x$ or

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

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

Hence, the correct answer is D.

Question 14: $\sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2}$ then $x$ is equal to

(A) $0,\frac{1}{2}$

(B) $1,\frac{1}{2}$

(C) 0

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

Answer:

Given the equation: $\sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2}$

we can migrate the $\sin^{-1}(1-x)$ term to the R.H.S.

Then we have;

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

or $- 2\sin^{-1}x =\cos^{-1}(1-x)$............................(1)

from $\left [\because \cos^{-1}(1-x) + \sin^{-1}(1-x) = \frac{\pi}{2} \right ]$

Take $\sin^{-1}x = \Theta$ $\Rightarrow \sin \Theta = x$ or $\cos \Theta = \sqrt{1-x^2}$ .

So, we conclude that;

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

Therefore we can put the value of $\sin^{-1}x$ in equation (1) we get,

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

Putting x = sin y in the above equation, we have then,

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

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

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

$\Rightarrow \cos(-2y) = 1-\sin y$

$\Rightarrow - 2y=\cos^{-1}(1-\sin y )$

$\Rightarrow 1- 2\sin^2 y = 1-\sin y$

$\Rightarrow 2\sin^2 y - \sin y = 0$

$\Rightarrow \sin y(2 \sin y -1) = 0$

So, we have the solution;

$\sin y = 0\ or\ \frac{1}{2}$ Therefore we have $x = 0\ or\ x= \frac{1}{2}$ .

When we have $x= \frac{1}{2}$ , we can see that :

$L.H.S. = \sin ^{-1}\left ( 1 - \frac{1}{2} \right ) - 2\sin^{-1}\frac{1}{2} = - \sin^{-1}\frac{1}{2} = -\frac{\pi}{6}$

So, it is not equal to the R.H.S. $-\frac{\pi}{6} \neq \frac{\pi}{2}$

Thus, we have only one solution, which is x = 0

Hence, the correct answer is (C).