Question

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

• A. $\sqrt{3}$
• B. $1$
• C. $-1$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is D $\frac{1}{\sqrt{3}}$

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

Let $x=\tan \theta$

$\Rightarrow {\tan }^{-1}\left(\frac{\tan \frac{\pi }{4}-\tan \theta }{1+\tan \frac{\pi }{4}\tan \theta }\right)=\frac{1}{2}{\tan }^{-1}\left(\tan \theta \right)$ $\because \tan \frac{\pi }{4}=1$

$\Rightarrow {\tan }^{-1}\tan (\frac{\pi }{4}-\theta )=\frac{1}{2}\theta$

$\Rightarrow \frac{\pi }{4}-\theta =\frac{1}{2}\theta$

$\Rightarrow \frac{\pi }{4}=\frac{3}{2}\theta$

$\Rightarrow \theta =\frac{\pi }{6}$

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