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:
$$\tan^{-1} \left(\dfrac {\sqrt {1+x^2}-\sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) =\alpha$$
$$\dfrac {\sqrt {1+x^2}-\sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}=\tan (\alpha )$$
$$\dfrac {1+x^2-2\sqrt {1+x^2}.\sqrt {1-x^2}+ 1 - x^2}{1+x^2-1+x^2}=\tan (\alpha)$$
$$\dfrac {1-\sqrt {1+x^2}.\sqrt {1-x^2}}{x^2}=\tan (\alpha)$$

Since
\begin{align}
\tan\alpha = \frac{\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}} =\frac{\text{opposite}}{\text{adjacent}}
\end{align}
then it follows
\begin{align}
\sin \alpha =& \frac{\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt{(\sqrt {1+x^2} - \sqrt {1-x^2})^2+(\sqrt {1+x^2} + \sqrt {1-x^2})^2}}\\
=&\ \frac{\sqrt {1+x^2} - \sqrt {1-x^2}}{2} = \frac{\text{opposite}}{\text{hypotenuse}}
\end{align}
and
\begin{align}
\cos\alpha = \frac{\sqrt {1+x^2} + \sqrt {1-x^2}}{2} = \frac{\text{adjacent}}{\text{hypotenuse}}.
\end{align}
Then it follows
\begin{align}
2\sin\alpha \cos \alpha = \frac{(\sqrt {1+x^2} + \sqrt {1-x^2})(\sqrt {1+x^2} - \sqrt {1-x^2})}{2}=x^2.
\end{align}
