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