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

<b>Case-I</b> <br/><br/> $-1 < x < 0$ <br/><br/> $\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\pi+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}$ <br/><br/> $\tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{-\pi}{3}$ <br/><br/> $2 \tan ^{-1} x=\frac{-\pi}{3}$ <br/><br/> $\tan ^{-1} x=\frac{-\pi}{6}$ <br/><br/> $x=\frac{-1}{\sqrt{3}}$ <br/><br/> <b>Case-II</b><br/><br/> $0 < x < 1$ <br/><br/> $\tan ^{-1} \frac{2 x}{1-x^{2}}+\tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{\pi}{3}$ <br/><br/> $$ \begin{aligned} & \tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{\pi}{6} \\\\ & 2 \tan ^{-1} x=\frac{\pi}{6} \\\\ & \tan ^{-1} x=\frac{\pi}{12} \\\\ & x=2-\sqrt{3} \\\\ & \text { Sum }=\frac{-1}{\sqrt{3}}+2-\sqrt{3}=2-\frac{4}{\sqrt{3}} \\\\ & \Rightarrow \alpha=2 \end{aligned} $$