Question

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

Answer 1

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

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

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

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

$\Rightarrow x=\tan \frac{\pi }{6}$

$\therefore x=\frac{1}{\sqrt{3}}$