If  ${\tan }^{-1}\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}=\alpha$, then  $x^2$ is equal to

The given equation can be written as  $\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}=\tan \alpha =\frac{\sin \alpha }{\cos \alpha }$

$\Rightarrow \frac{2\sqrt{1+x^2}}{2\sqrt{1-x^2}}=\frac{\cos \alpha +\sin \alpha }{\cos \alpha -\sin \alpha }$

[Applying componendo and dividendo]

$\Rightarrow \frac{1+x^2}{1-x^2}=\frac{1+\sin 2\alpha }{1-\sin 2\alpha }$ 

$\Rightarrow \frac{1-x^2}{1+x^2}=\frac{1-\sin 2\alpha }{1+\sin 2\alpha }$

Squaring on both sides

$\frac{1-x^2+1+x^2}{1-x^2-1-x^2}=\frac{1-\sin 2\alpha +1+\sin 2\alpha }{1-\sin 2\alpha -1-\sin 2\alpha }$

$\frac{2}{-2x^2}=\frac{2}{-2\sin 2\alpha }$

$x^2=\sin 2\alpha$.