If $\tan^{-1} \left(\frac {\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) = \alpha$, then prove $x^2=\sin(2\alpha)$ If $\tan^{-1} \left(\dfrac {\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right)  = \alpha$ then prove that: $x^2= \sin (2\alpha) $
My Attempt:
$$\tan^{-1} \left(\dfrac {\sqrt {1+x^2}-\sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) =\alpha$$
$$\dfrac {\sqrt {1+x^2}-\sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}=\tan (\alpha )$$
$$\dfrac {1+x^2-2\sqrt {1+x^2}.\sqrt {1-x^2}+ 1 - x^2}{1+x^2-1+x^2}=\tan (\alpha)$$
$$\dfrac {1-\sqrt {1+x^2}.\sqrt {1-x^2}}{x^2}=\tan (\alpha)$$

Let $A=\sqrt{1+x^2}$ and $B=\sqrt{1-x^2}$, then
$$\tan(\alpha) = \frac{A-B}{A+B}, \quad A^2+B^2=2,\quad A^2-B^2 = 2x^2$$
and also note
$$\tan(a) = \underbrace{\frac{\sin(2a)}{2\sin(a)\cos(a)}}_{=\,1} \cdot \frac{\sin(a)}{\cos(a)}= \sin(2a)\cdot\frac{1}{2}\sec^2(a) = \sin(2a) \frac{1}{2} \left(1+\tan^2(a)\frac{}{}\right)$$
hence,
\begin{align}
\sin(2a)
&= \frac{2 \tan(a)}{1+\tan^2(a)}
= \frac{2 \tan(a)}{1+\tan^2(a)} \cdot \frac{\big(A+B\big)^2}{\big(A+B\big)^2}\\
&= \frac{2 (A+B)(A-B)}{(A+B)^2 + (A-B)^2}
= \frac{2 (A^2-B^2)}{2A^2+2B^2} = x^2
\end{align}
