Question

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

Answer 1

tan-1$\frac {1-x}{1+x}$ = $\frac12$ tan-1x
tan-11 - tan-1x = $\frac 12$ tan-1x
\implies$$\frac {\pi}{4} = $\frac 32$ tan-1x
$\implies$tan-1x = $\frac {\pi}{6}$
$\implies$x = tan-1$\frac {\pi}{6}$
So, x = $\frac {1}{\sqrt 3}$