Question

Find the real solutions of the equation ${\tan }^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}{\tan }^{-1}x,(x>0)$.

Answer 1

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

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

$2[ta{n}^{-1}1-ta{n}^{-1}x]=ta{n}^{-1}x$

$2\times \frac{\pi }{4}=3ta{n}^{-1}x$

$ta{n}^{-1}x=\frac{\pi }{6}$

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