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

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

Answer:

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

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

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

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

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

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

$\frac{\pi }{6}ϵ[0,\pi ]$

Therefore we have,

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

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

Answer:

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

so, as we know ${\tan }^{-1}(\tan x)=xifxϵ(-\frac{\pi }{2},\frac{\pi }{2})$

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

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

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

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

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

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

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

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

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

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 $[2{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2}]$

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

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

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 similarly;

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

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,

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

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

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

$={\tan }^{-1}\left(\frac{56}{33}\right)$ 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 in 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 $[{\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy}]$

so,

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

equal to L.H.S

Hence proved.

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

Answer:

Applying the formlua:

${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy}$ on two parts.

we will have,

$={\tan }^{-1}\left(\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{5}\times \frac{1}{7}}\right)+{\tan }^{-1}\left(\frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{3}\times \frac{1}{8}}\right)$

$={\tan }^{-1}\left(\frac{7+5}{35-1}\right)+{\tan }^{-1}\left(\frac{8+3}{24-1}\right)$

$={\tan }^{-1}\left(\frac{12}{34}\right)+{\tan }^{-1}\left(\frac{11}{23}\right)$

$={\tan }^{-1}\left(\frac{6}{17}\right)+{\tan }^{-1}\left(\frac{11}{23}\right)$

$={\tan }^{-1}\left[\frac{\frac{6}{17}+\frac{11}{23}}{1-\frac{6}{17}\times \frac{11}{23}}\right]$

$={\tan }^{-1}\left[\frac{325}{325}\right]={\tan }^{-1}1$

$=\frac{\pi }{4}$

Hence it s equal to R.H.S

Proved.

Question:9 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:10 Prove that ${\cot }^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2},x\in (0,\frac{\pi }{4})$

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 rationalize the fraction

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

We get then,

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

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

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

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

Therefore we can write it as;

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

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

Hence proved.

Question:11 Prove that ${\tan }^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}x,-\frac{1}{\sqrt{2}}\le x\le 1$

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

Answer:

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

we get then,

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

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

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

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

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

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

Hence proved

Question:12 Prove that $\frac{9\pi }{8}-\frac{9}{4}{\sin }^{-1}\frac{1}{3}=\frac{9}{4}{\sin }^{-1}\frac{2\sqrt{2}}{3}$

Answer:

We have to solve the given equation:

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

Take $\frac{9}{4}$ as common in L.H.S,

$=\frac{9}{4}[\frac{\pi }{2}-{\sin }^{-1}\frac{1}{3}]$

or $=\frac{9}{4}[{\cos }^{-1}\frac{1}{3}]$ from $[{\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}]$

Now, assume,

$[{\cos }^{-1}\frac{1}{3}]=y$

Then,

$\cos y=\frac{1}{3}\Rightarrow \sin y=\sqrt{1-(\frac{1}{3}{)}^2}=\frac{2.\sqrt{2}}{3}$

Therefore we have now,

$y={\sin }^{-1}\frac{2.\sqrt{2}}{3}$

So we have L.H.S then $=\frac{9}{4}{\sin }^{-1}\frac{2.\sqrt{2}}{3}$

That is equal to R.H.S.

Hence proved.

Question:13 Solve the following equations: $2{\tan }^{-1}(\cos x)={\tan }^{-1}(2\text{cosec}x)$

Answer:

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

Using the formula:

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

We can write

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

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

So, we can equate;

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

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

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

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

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

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

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

$\Rightarrow x=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$

Hence the value of $x=\frac{1}{\sqrt{3}}$ .

Question:15 $\sin ({\tan }^{-1}x),|x|<1$ 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

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

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

Hence the correct answer is D.

Question:16 ${\sin }^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2}$ 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 $[\because {\cos }^{-1}(1-x)+{\sin }^{-1}(1-x)=\frac{\pi }{2}]$

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

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,

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

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

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

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

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

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

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

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

So, we have the solution;

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

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

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

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

Thus we have only one solution which is x = 0

Hence the correct answer is (C).

Question:17 ${\tan }^{-1}\left(\frac{x}{y}\right)-{\tan }^{-1}\frac{x-y}{x+y}$ is equal to

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

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

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

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

Answer:

Applying formula: $[{\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+xy}\right)]$ .

We get,

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

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

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

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

Hence, the correct answer is C.