If \(\tan ^{-1}\left[\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right]=\alpha\), then the value of \(\sin 2 \alpha\) is

We have \(\tan \alpha=\left[\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right]\)
Put \(x^2=\cos \theta\)
\(\therefore \tan \alpha=\frac{\sqrt{2 \cos ^2 \frac{\theta}{2}-\sqrt{2 \sin ^2 \frac{\theta}{2}}}}{\sqrt{2 \cos ^2 \frac{\theta}{2}+\sqrt{2 \sin ^2 \frac{\theta}{2}}}} \)
\( \therefore \tan \alpha=\frac{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}=\frac{1-\tan \frac{\theta}{2}}{1+\tan \frac{\theta}{2}}=\) \(\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right) \)
\( \therefore \alpha=\frac{\pi}{4}-\frac{\theta}{2} \Rightarrow 2 \alpha=\frac{\pi}{2}-\theta \)
\( \therefore \sin 2 \alpha=\sin \left(\frac{\pi}{2}-\theta\right)=\cos \theta=x^2\)