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 = \(\frac {1}{\sqrt 3}\)