# Miscellaneous Exercise Inverse Trigonometric Functions Solution - NCERT Class 12

Go back to  'Inverse Trigonometric Functions'

## Chapter 2 Ex.2.ME Question 1

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

### Solution

\begin{align}{\cos ^{ - 1}}\left( {\cos \frac{{13\pi }}{6}} \right)& = {\cos ^{ - 1}}\left[ {\cos \left( {2\pi + \frac{\pi }{6}} \right)} \right]\\&= {\cos ^{ - 1}}\left[ {\cos \frac{\pi }{6}} \right]\\&= \frac{\pi }{6}\end{align}

## Chapter 2 Ex.2.ME Question 2

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

### Solution

\begin{align}{\tan ^{ - 1}}\left( {\tan \frac{{7\pi }}{6}} \right) &= {\tan ^{ - 1}}\left[ {\tan \left( {2\pi - \frac{{5\pi }}{6}} \right)} \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 \frac{\pi }{6}} \right]\\&= \frac{\pi }{6}\end{align}

## Chapter 2 Ex.2.ME Question 3

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

### Solution

Let $${\sin ^{ - 1}}\frac{3}{5} = x \Rightarrow \sin x = \frac{3}{5}$$

Then,

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

Therefore,

\begin{align}\tan x &= \frac{3}{4}\\x &= {\tan ^{ - 1}}\frac{3}{4}\\{\sin ^{ - 1}}\frac{3}{5} &= {\tan ^{ - 1}}\frac{3}{4}\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Thus,

\begin{align}LHS &= 2{\sin ^{ - 1}}\frac{3}{5}\\&= 2{\tan ^{ - 1}}\frac{3}{4}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {from\;\left( 1 \right)} \right]\\&= {\tan ^{ - 1}}\left( {\frac{{2 \times \frac{3}{4}}}{{1 - {{\left( {\frac{3}{4}} \right)}^2}}}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{24}}{7}} \right)\\&= RHS\end{align}

## Chapter 2 Ex.2.ME Question 4

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

### Solution

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

Then,

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

Therefore,

\begin{align}\tan x &= \frac{8}{{15}}\\x &= {\tan ^{ - 1}}\frac{8}{{15}}\\{\sin ^{ - 1}}\frac{8}{{17}} &= {\tan ^{ - 1}}\frac{8}{{15}}\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

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

Then,

$$\cos y = \sqrt {1 - {{\left( {\frac{3}{5}} \right)}^2}} = \sqrt {\frac{{16}}{{25}}} = \frac{4}{5}$$

Therefore,

\begin{align}\tan y &= \frac{3}{4}\\y &= {\tan ^{ - 1}}\frac{3}{4}\\{\sin ^{ - 1}}\frac{3}{5}& = {\tan ^{ - 1}}\frac{3}{4}\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Thus, by using $$\left( 1 \right)$$ and $$\left( 2 \right)$$

\begin{align}LHS &= {\sin ^{ - 1}}\frac{8}{{17}} + {\sin ^{ - 1}}\frac{3}{5}\\&= {\tan ^{ - 1}}\frac{8}{{15}} + {\tan ^{ - 1}}\frac{3}{4}\\&= {\tan ^{ - 1}}\left[ {\frac{{\frac{8}{{15}} + \frac{3}{4}}}{{1 - \frac{8}{{15}} \cdot \frac{3}{4}}}} \right]\\&= {\tan ^{ - 1}}\left[ {\frac{{\frac{{32 + 45}}{{60}}}}{{\frac{{60 - 24}}{{60}}}}} \right]\\&= {\tan ^{ - 1}}\frac{{77}}{{36}}\\&= RHS\end{align}

## Chapter 2 Ex.2.ME Question 5

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

### Solution

Let $${\cos ^{ - 1}}\frac{4}{5} = x \Rightarrow \cos x = \frac{4}{5}$$

Then,

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

Therefore,

\begin{align}\tan x &= \frac{3}{4}\\x &= {\tan ^{ - 1}}\frac{3}{4}\\{\cos ^{ - 1}}\frac{4}{5} &= {\tan ^{ - 1}}\frac{3}{4}\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

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

Then,

$$\sin y = \frac{5}{{13}}$$

Therefore,

\begin{align}\tan y &= \frac{5}{{12}}\\y &= {\tan ^{ - 1}}\frac{5}{{12}}\\{\cos ^{ - 1}}\frac{{12}}{{13}} &= {\tan ^{ - 1}}\frac{5}{{12}}\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Thus, by using $$\left( 1 \right)$$ and $$\left( 2 \right)$$

\begin{align}{\cos ^{ - 1}}\frac{4}{5} + {\cos ^{ - 1}}\frac{{12}}{{13}}& = {\tan ^{ - 1}}\frac{3}{4} + {\tan ^{ - 1}}\frac{5}{{12}}\\&= {\tan ^{ - 1}}\left[ {\frac{{\frac{3}{4} + \frac{5}{{12}}}}{{1 - \frac{3}{4} \cdot \frac{5}{{12}}}}} \right]\\&= {\tan ^{ - 1}}\left[ {\frac{{56}}{{33}}} \right]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}

Now, let $${\cos ^{ - 1}}\frac{{33}}{{65}} = z \Rightarrow \cos z = \frac{{33}}{{65}}$$

Then,

$$\sin z = \sqrt {1 - {{\left( {\frac{{33}}{{65}}} \right)}^2}} = \frac{{56}}{{65}}$$

Therefore,

\begin{align}\tan z &= \frac{{33}}{{56}}\\z &= {\tan ^{ - 1}}\frac{{56}}{{33}}\\{\cos ^{ - 1}}\frac{{33}}{{65}} &= {\tan ^{ - 1}}\frac{{56}}{{33}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 4 \right)\end{align}

Thus, by using $$\left( 3 \right)$$ and $$\left( 4 \right)$$

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

Hence proved.

## Chapter 2 Ex.2.ME Question 6

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

### Solution

Let $${\cos ^{ - 1}}\frac{{12}}{{13}} = y \Rightarrow \cos y = \frac{{12}}{{13}}$$

Then,

$$\sin y = \sqrt {1 - {{\left( {\frac{{12}}{{13}}} \right)}^2}} = \frac{5}{{13}}$$

Therefore,

\begin{align}\tan y &= \frac{5}{{12}}\\y &= {\tan ^{ - 1}}\frac{5}{{12}}\\{\cos ^{ - 1}}\frac{{12}}{{13}} &= {\tan ^{ - 1}}\frac{5}{{12}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Now, let $${\sin ^{ - 1}}\frac{3}{5} = x \Rightarrow \sin x = \frac{3}{5}$$

Then,

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

Therefore,

\begin{align}\tan x &= \frac{3}{4}\\x &= {\tan ^{ - 1}}\frac{3}{4}\\{\sin ^{ - 1}}\frac{3}{5} &= {\tan ^{ - 1}}\frac{3}{4}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Now, let $${\sin ^{ - 1}}\frac{{56}}{{65}} = z \Rightarrow \sin z = \frac{{56}}{{65}}$$

Then,

$$\cos z = \sqrt {1 - {{\left( {\frac{{56}}{{65}}} \right)}^2}} = \frac{{33}}{{65}}$$

Therefore,

\begin{align}\tan z &= \frac{{56}}{{33}}\\z &= {\tan ^{ - 1}}\frac{{56}}{{33}}\\{\sin ^{ - 1}}\frac{{56}}{{65}} &= {\tan ^{ - 1}}\frac{{56}}{{33}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}

Thus, by using $$\left( 1 \right)$$ and $$\left( 2 \right)$$

\begin{align}LHS& = {\cos ^{ - 1}}\frac{{12}}{{13}} + {\sin ^{ - 1}}\frac{3}{5}\\&= {\tan ^{ - 1}}\frac{5}{{12}} + {\tan ^{ - 1}}\frac{3}{4}\\&= {\tan ^{ - 1}}\left[ {\frac{{\frac{5}{{12}} + \frac{3}{4}}}{{1 - \frac{5}{{12}} \cdot \frac{3}{4}}}} \right]\\&= {\tan ^{ - 1}}\left[ {\frac{{\frac{{20 + 36}}{{48}}}}{{\frac{{48 - 15}}{{48}}}}} \right]\\&= {\tan ^{ - 1}}\left( {\frac{{56}}{{33}}} \right)\\&= {\sin ^{ - 1}}\frac{{56}}{{65}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{Using}}\left( {\rm{3}} \right)} \right]\\&= RHS\end{align}

## Chapter 2 Ex.2.ME Question 7

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

### Solution

Let $${\sin ^{ - 1}}\frac{5}{{13}} = x \Rightarrow \sin x = \frac{5}{{13}}$$

Then,

$$\cos x = \sqrt {1 - {{\left( {\frac{5}{{13}}} \right)}^2}} = \frac{{12}}{{13}}$$

Therefore,

\begin{align}\tan x&= \frac{5}{{12}}\\x &= {\tan ^{ - 1}}\frac{5}{{12}}\\{\sin ^{ - 1}}\frac{5}{{13}} &= {\tan ^{ - 1}}\frac{5}{{12}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

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

Then,

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

Therefore,

\begin{align}\tan y &= \frac{4}{3}\\y &= {\tan ^{ - 1}}\frac{4}{3}\\{\cos ^{ - 1}}\frac{3}{5} &= {\tan ^{ - 1}}\frac{4}{3}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Thus, by using $$\left( 1 \right)$$ and $$\left( 2 \right)$$

\begin{align}RHS&= {\sin ^{ - 1}}\frac{5}{{12}} + {\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}} \cdot \frac{4}{3}}}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{63}}{{16}}} \right)\\&= LHS\end{align}

## Chapter 2 Ex.2.ME Question 8

Prove that $${\tan ^{ - 1}}\frac{1}{5} + {\tan ^{ - 1}}\frac{1}{7} + {\tan ^{ - 1}}\frac{1}{3} + {\tan ^{ - 1}}\frac{1}{8} = \frac{\pi }{4}$$

### Solution

\begin{align}LHS&= {\tan ^{ - 1}}\frac{1}{5} + {\tan ^{ - 1}}\frac{1}{7} + {\tan ^{ - 1}}\frac{1}{3} + {\tan ^{ - 1}}\frac{1}{8}\\&= {\tan ^{ - 1}}\left( {\frac{{\frac{1}{5} + \frac{1}{7}}}{{1 - \frac{1}{5} \cdot \frac{1}{7}}}} \right) + {\tan ^{ - 1}}\left( {\frac{{\frac{1}{3} + \frac{1}{8}}}{{1 - \frac{1}{3} \cdot \frac{1}{8}}}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{12}}{{34}}} \right) + {\tan ^{ - 1}}\left( {\frac{{11}}{{23}}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{6}{{17}}} \right) + {\tan ^{ - 1}}\left( {\frac{{11}}{{23}}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{\frac{6}{{17}} + \frac{{11}}{{23}}}}{{1 - \frac{6}{{17}} \cdot \frac{{11}}{{23}}}}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{325}}{{325}}} \right)\\&= {\tan ^{ - 1}}\left( 1 \right)\\&= \frac{\pi }{4}\\&= RHS\end{align}

## Chapter 2 Ex.2.ME Question 9

Prove that $${\tan ^{ - 1}}\sqrt x = \frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right),x \in \left[ {0,1} \right]$$.

### Solution

Let $$x = {\tan ^2}\theta$$

Then,

\begin{align}\sqrt x &= \tan \theta \\\theta &= {\tan ^{ - 1}}\sqrt x \end{align}

Therefore,

\begin{align}\left( {\frac{{1 - x}}{{1 + x}}} \right) &= \frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}\\&= \cos 2\theta \end{align}

Thus,

\begin{align}RHS &= \frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right)\\&= \frac{1}{2}{\cos ^{ - 1}}\left( {\cos 2\theta } \right)\\&= \frac{1}{2} \times 2\theta \\&= \theta \\&= {\tan ^{ - 1}}\sqrt x \\&= LHS\end{align}

## Chapter 2 Ex.2.ME Question 10

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

### Solution

\begin{align}\left( {\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right) &= \frac{{{{\left( {\sqrt {1 + \sin x} + \sqrt {1 - \sin x} } \right)}^2}}}{{{{\left( {\sqrt {1 + \sin x} } \right)}^2} - {{\left( {\sqrt {1 - \sin x} } \right)}^2}}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {\rm{by}\,{\rm{ rationalizing}}} \right)\\&= \frac{{\left( {1 + \sin x} \right) + \left( {1 - \sin x} \right) + 2\sqrt {\left( {1 + \sin x} \right)\left( {1 - \sin x} \right)} }}{{1 + \sin x - 1 + \sin x}}\\&= \frac{{2\left( {1 + \sqrt {1 - {{\sin }^2}x} } \right)}}{{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}\end{align}

Thus,

\begin{align}LHS &= {\cot ^{ - 1}}\left( {\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right)\\&= {\cot ^{ - 1}}\left( {\cot \frac{x}{2}} \right)\\&= \frac{x}{2}\\&= RHS\end{align}

## Chapter 2 Ex.2.ME Question 11

Prove that $${\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{\sqrt {1 + x} + \sqrt {1 - x} }}} \right) = \frac{\pi }{2} - \frac{1}{2}{\cos ^{ - 1}}x, - \frac{1}{{\sqrt 2 }} \le x \le 1$$

### Solution

Let $$x = \cos 2\theta \Rightarrow \theta = \frac{1}{2}{\cos ^{ - 1}}x$$

Thus,

\begin{align}LHS &= {\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{\sqrt {1 + x} + \sqrt {1 - x} }}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + \cos 2\theta } - \sqrt {1 - \cos 2\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{{\sqrt 2 \cos \theta - \sqrt 2 \sin \theta }}{{\sqrt 2 \cos \theta + \sqrt 2 \sin \theta }}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{\cos \theta - \sin \theta }}{{\cos \theta + \sin \theta }}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{1 - \tan \theta }}{{1 + \tan \theta }}} \right)\\&= {\tan ^{ - 1}}1 - {\tan ^{ - 1}}\left( {\tan \theta } \right)\\&= \frac{\pi }{4} - \theta \\&= \frac{\pi }{4} - \frac{1}{2}{\cos ^{ - 1}}x\\&= RHS\end{align}

## Chapter 2 Ex.2.ME Question 12

Prove that $$\frac{{9\pi }}{8} - \frac{9}{4}{\sin ^{ - 1}}\frac{1}{3} = \frac{9}{4}{\sin ^{ - 1}}\frac{{2\sqrt 2 }}{3}$$

### Solution

\begin{align}LHS &= \frac{{9\pi }}{8} - \frac{9}{4}{\sin ^{ - 1}}\frac{1}{3}\\&= \frac{9}{4}\left( {\frac{\pi }{2} - {{\sin }^{ - 1}}\frac{1}{3}} \right)\\&= \frac{9}{4}\left( {{{\cos }^{ - 1}}\frac{1}{3}} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Now, let $${\cos ^{ - 1}}\frac{1}{3} = x \Rightarrow \cos x = \frac{1}{3}$$

Therefore,

\begin{align}\sin x& = \sqrt {1 - {{\left( {\frac{1}{3}} \right)}^2}} \\&= \frac{{2\sqrt 2 }}{3}\\x &= {\sin ^{ - 1}}\frac{{2\sqrt 2 }}{3}\\{\cos ^{ - 1}}\frac{1}{3} &= {\sin ^{ - 1}}\frac{{2\sqrt 2 }}{3}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Thus, by using $$\left( 1 \right)$$ and $$\left( 2 \right)$$

$\frac{{9\pi }}{8} - \frac{9}{4}{\sin ^{ - 1}}\frac{1}{3} = \frac{9}{4}{\sin ^{ - 1}}\frac{{2\sqrt 2 }}{3}$

Hence proved.

## Chapter 2 Ex.2.ME Question 13

Solve $$2{\tan ^{ - 1}}\left( {\cos x} \right) = {\tan ^{ - 1}}\left( {2{\rm{cosec }}x} \right)$$.

### Solution

It is given that $$2{\tan ^{ - 1}}\left( {\cos x} \right) = {\tan ^{ - 1}}\left( {2{\rm{cosec }}x} \right)$$

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

Hence,$${\tan ^{ - 1}}\frac{{1 - x}}{{1 + x}} = \frac{1}{2}{\tan ^{ - 1}}x,\left( {x > 0} \right)$$

\begin{align}&\Rightarrow {\tan ^{ - 1}}\left( {\frac{{2\cos x}}{{1 - {{\cos }^2}x}}} \right) = {\tan ^{ - 1}}\left( {2{\rm{cosec }}x} \right)\\&\Rightarrow \left( {\frac{{2\cos x}}{{1 - {{\cos }^2}x}}} \right) = \left( {2{\rm{cosec }}x} \right)\\&\Rightarrow \frac{{2\cos x}}{{{{\sin }^2}x}} = \frac{2}{{\sin x}}\\&\Rightarrow \cos x = \sin x\\&\Rightarrow \tan x = 1\\&\Rightarrow \tan x = \tan \frac{\pi }{4}\end{align}

Therefore,

$$x = n\pi + \frac{\pi }{4}$$, where $$n \in Z$$.

## Chapter 2 Ex.2.ME Question 14

Solve $${\tan ^{ - 1}}\frac{{1 - x}}{{1 + x}} = \frac{1}{2}{\tan ^{ - 1}}x,\left( {x > 0} \right)$$

### Solution

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

Hence,

\begin{align}&\Rightarrow {\tan ^{ - 1}}\frac{{1 - x}}{{1 + x}} = \frac{1}{2}{\tan ^{ - 1}}x\\&\Rightarrow {\tan ^{ - 1}}1 - {\tan ^{ - 1}}x = \frac{1}{2}{\tan ^{ - 1}}x\\&\Rightarrow \frac{\pi }{4} = \frac{3}{2}{\tan ^{ - 1}}x\\&\Rightarrow x = \tan \frac{\pi }{6}\\&z\Rightarrow x = \frac{1}{{\sqrt 3 }}\end{align}

## Chapter 2 Ex.2.ME Question 15

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

### Solution

Let $$\tan y = x$$

Therefore,

$$\sin y = \frac{x}{{\sqrt {1 + {x^2}} }}$$

Now, let $${\tan ^{ - 1}}x = y$$

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

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

Hence,

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

Thus,

\begin{align}\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}} }}\end{align}

Thus, the correct option is D.

## Chapter 2 Ex.2.ME Question 16

Solve: $${\sin ^{ - 1}}\left( {1 - x} \right) - 2{\sin ^{ - 1}}x = \frac{\pi }{2}$$, then is equal to

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

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

(C) $$0$$

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

### Solution

It is given that $${\sin ^{ - 1}}\left( {1 - x} \right) - 2{\sin ^{ - 1}}x = \frac{\pi }{2}$$

\begin{align}&\Rightarrow {\sin ^{ - 1}}\left( {1 - x} \right) - 2{\sin ^{ - 1}}x = \frac{\pi }{2}\\&\Rightarrow - 2{\sin ^{ - 1}}x = \frac{\pi }{2} - {\sin ^{ - 1}}\left( {1 - x} \right)\\&\Rightarrow - 2{\sin ^{ - 1}}x = {\cos ^{ - 1}}\left( {1 - x} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Let $${\sin ^{ - 1}}x = y \Rightarrow \sin y = x$$

Hence,

\begin{align}&\cos y = \sqrt {1 - {x^2}} \\&y = {\cos ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right)\\&{\sin ^{ - 1}}x = {\cos ^{ - 1}}\sqrt {1 - {x^2}} \end{align}

From equation (1), we have

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

Put $$x = \sin y$$

\begin{align}&\Rightarrow - 2{\cos ^{ - 1}}\sqrt {1 - {{\sin }^2}y} = {\cos ^{ - 1}}\left( {1 - \sin y} \right)\\&\Rightarrow - 2{\cos ^{ - 1}}\left( {\cos y} \right) = {\cos ^{ - 1}}\left( {1 - \sin y} \right)\\&\Rightarrow - 2y = {\cos ^{ - 1}}\left( {1 - \sin y} \right)\\&\Rightarrow 1 - \sin y = \cos \left( { - 2y} \right)\\&\Rightarrow 1 - \sin y = \cos 2y\\&\Rightarrow 1 - \sin y = 1 - 2{\sin ^2}y\\&\Rightarrow 2{\sin ^2}y - \sin y = 0\\&\Rightarrow \sin y\left( {2\sin y - 1} \right) = 0\\&\Rightarrow \sin y = 0,\frac{1}{2}\end{align}

Therefore,

$$x = 0,\frac{1}{2}$$

When $$x = \frac{1}{2}$$, it does not satisfy the equation.

Hence, $$x = 0$$ is the only solution

Thus, the correct option is C.

## Chapter 2 Ex.2.ME Question 17

Solve $${\tan ^{ - 1}}\left( {\frac{x}{y}} \right) - {\tan ^{ - 1}}\frac{{x - y}}{{x + y}}$$ is equal to

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

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

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

(D) $$\frac{{ - 3\pi }}{4}$$

### Solution

\begin{align}{\tan ^{ - 1}}\left( {\frac{x}{y}} \right) - {\tan ^{ - 1}}\frac{{x - y}}{{x + y}}& = {\tan ^{ - 1}}\left[ {\frac{{\frac{x}{y} - \frac{{x - y}}{{x + y}}}}{{1 + \left( {\frac{x}{y}} \right)\left( {\frac{{x - y}}{{x + y}}} \right)}}} \right]\\&= {\tan ^{ - 1}}\left[ {\frac{{\frac{{x\left( {x + y} \right) - y\left( {x - y} \right)}}{{y\left( {x + y} \right)}}}}{{\frac{{y\left( {x + y} \right) + x\left( {x - y} \right)}}{{y\left( {x + y} \right)}}}}} \right]\\&= {\tan ^{ - 1}}\left( {\frac{{{x^2} + xy - xy + {y^2}}}{{xy + {y^2} + {x^2} - xy}}} \right)\\&= {\tan ^{ - 1}}\left( 1 \right)\\&= {\tan ^{ - 1}}\left( {\tan \frac{\pi }{4}} \right)\\&= \frac{\pi }{4}\end{align}

Thus, the correct option is $$C$$.

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