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 $\alpha =$ 

$x>0{\cot }^{-1}x={\tan }^{-1}\frac{1}{x}$
$\Rightarrow 2{\tan }^{-1}x+2{\tan }^{-1}x=\frac{\pi }{3}$
$\Rightarrow 4{\tan }^{-1}x=\frac{\pi }{3}$
$\Rightarrow {\tan }^{-1}x=\frac{\pi }{12}$
$x=\tan \frac{\pi }{12}=2-\sqrt{3}$
$x<0\Rightarrow {\cot }^{-1}x=\pi +{\tan }^{-1}\frac{1}{x}$
$2{\tan }^{-1}x+\pi +2{\tan }^{-1}x=\frac{\pi }{3}$
$4{\tan }^{-1}x=\frac{-2\pi }{3}$
${\tan }^{-1}x=\frac{-\pi }{6}$
$x=\frac{-1}{\sqrt{3}}$
Sum of value of 
$x=2-\sqrt{3}-\frac{1}{\sqrt{3}}$
$=2-\frac{4}{\sqrt{3}}$
$\alpha =2$