Question

Solve for x: ${\tan }^{-1}\left[\frac{1-x}{1+x}\right]-\frac{1}{2}{\tan }^{-1}x=0$ , x>0

Answer 1

${\tan }^{-1}\left[\frac{1-x}{1+x}\right]-\frac{1}{2}{\tan }^{-1}x=0$

Let $x=\tan \theta$

$={\tan }^{-1}\left(\frac{1-\tan \theta }{1+\tan \theta }\right)-\frac{1}{2}{\tan }^{-1}x=0$

$={\tan }^{-1}\frac{\tan \frac{\pi }{4}-\tan \theta }{1+\tan \frac{\pi }{4}\tan \theta }-\frac{1}{2}{\tan }^{-1}x=0$

$\Rightarrow {\tan }^{-1}\left(\tan \right(\frac{\pi }{4}-\theta )-\frac{1}{2}{\tan }^{-1}x=0$

$\Rightarrow \frac{\pi }{4}-{\tan }^{-1}x-\frac{1}{2}{\tan }^{-1}x=\frac{\pi }{4}-\frac{3}{2}{\tan }^{-1}x=0$

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

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

$\Rightarrow x=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$