Question

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

Answer 1

${\tan }^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}{\tan }^{-1}x$
${\tan }^{-1}\left(1\right)-{\tan }^{-1}x=\frac{1}{2}{\tan }^{-1}x$
Let ${\tan }^{-1}x=\theta$
$\frac{\pi }{4}-\theta =\frac{\theta }{2}\Rightarrow \theta +\frac{\theta }{2}=\frac{\pi }{4}$
$3\frac{\theta }{2}=\frac{\pi }{4}$
$\theta =\frac{\pi }{6}$
$x=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$