<s>[INST] 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:
$$\tan^{-1} \left(\dfrac {\sqrt {1+x^2}-\sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) =\alpha$$
$$\dfrac {\sqrt {1+x^2}-\sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}=\tan (\alpha )$$
$$\dfrac {1+x^2-2\sqrt {1+x^2}.\sqrt {1-x^2}+ 1 - x^2}{1+x^2-1+x^2}=\tan (\alpha)$$
$$\dfrac {1-\sqrt {1+x^2}.\sqrt {1-x^2}}{x^2}=\tan (\alpha)$$
 [/INST] As for real $\alpha, x^2\le1$
WLOG $x^2=\cos2y\implies0\le2y\le\dfrac\pi2\implies\cos y,\sin y\ge0$
Using $\cos2y=1-2\sin^2y=2\cos^2y-1,$
$$\tan\alpha=\dfrac{\cos y-\sin y}{\cos y+\sin y}=\tan\left(\dfrac\pi4-y\right)$$
$\implies\alpha=m\pi+\dfrac\pi4-y$ where $m$ is any integer
$x^2=\cos2y=\cos\left(2m\pi+\dfrac\pi2-2\alpha\right)=?$
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