Question

If the sum of all the solutions of ${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+{\cot }^{-1}\left(\frac{1-x^2}{2x}\right)=\frac{\pi }{3},-1<x<1,x\ne 0$, is $\alpha -\frac{4}{\sqrt{3}}$, then $\alpha$ is equal to ____

Answer 1

Answer: 2

Case $I:x>0$
${\tan }^{-1}\frac{2x}{1-x^2}+{\tan }^{-1}\frac{2x}{1-x^2}=\frac{\pi }{3}$
$x=2-\sqrt{3}$
Case II : $x<0$
${\tan }^{-1}\frac{2x}{1-x^2}+{\tan }^{-1}\frac{2x}{1-x^2}+\pi =\frac{\pi }{3}$
$x=\frac{-1}{\sqrt{3}}\Rightarrow \alpha =2$