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 $\frac{1}{\sqrt{3}}$

${\tan }^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}{\tan }^{-1}x,x>0\cdots \left(1\right)$
$\text{Put}x=\tan \theta ,\theta \in (0,\frac{\pi }{2})$
Then $\left(1\right)$ becomes
${\tan }^{-1}\left(\tan (\frac{\pi }{4}-\theta )\right)=\frac{1}{2}{\tan }^{-1}\left(\tan \theta \right)$
$\Rightarrow \frac{\pi }{4}-\theta =\frac{\theta }{2}$
$\Rightarrow \theta =\frac{\pi }{6}$
$\Rightarrow {\tan }^{-1}x=\frac{\pi }{6}$
$\Rightarrow x=\frac{1}{\sqrt{3}}$