<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] $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note that
  \begin{align}
\pars{{p \over q} = z \implies
\left\{\begin{array}{lcl}
\ds{p + q \over q} & \ds{=} & \ds{z + 1}
\\[1mm]
\mbox{and}&&
\\[1mm]
\ds{p - q \over q} & \ds{=} & \ds{z - 1}
\end{array}\right.}
\implies \bbx{{p \over q} = z \implies {p + q \over p - q} =
{z + 1 \over z - 1}}\label{1}\tag{1}
\end{align}

Then,
\begin{align}
&\arctan\pars{\root{1 + x^{2}} - \root{1 - x^{2}} \over
\root{1 + x^{2}} + \root{1 - x^{2}}}  = \alpha
\implies
{\root{1 + x^{2}} - \root{1 - x^{2}} \over
\root{1 + x^{2}} + \root{1 - x^{2}}}  = \tan\pars{\alpha}
\end{align}

With the identity \eqref{1}:

\begin{align}
&{2\root{1 + x^{2}} \over
-2\root{1 - x^{2}}}  = {\tan\pars{\alpha} + 1 \over \tan\pars{\alpha} - 1}
\implies 
{1 + x^{2} \over 1 - x^{2}} = \bracks{\tan\pars{\alpha} + 1 \over \tan\pars{\alpha} - 1}^{2}
\\[5mm] \stackrel{\mrm{see}\ \eqref{1}}{\implies}\,\,\ &
{2 \over 2x^{2}} = {\braces{\bracks{\tan\pars{\alpha} + 1}/
\bracks{\tan\pars{\alpha} - 1}}^{\,2} + 1 \over
\braces{\bracks{\tan\pars{\alpha} + 1}/
\bracks{\tan\pars{\alpha} - 1}}^{\,2} - 1} =
{2\tan^{2}\pars{\alpha} + 2 \over 4\tan\pars{\alpha}}
\\[5mm] \implies &\
x^{2} =
{2\tan\pars{\alpha} \over \tan^{2}\pars{\alpha} + 1} =
{2\tan\pars{\alpha} \over\sec^{2}\pars{\alpha}} =
2\sin\pars{\alpha}\cos\pars{\alpha} \implies \bbx{x^{2} = \sin\pars{2\alpha}}  
\end{align}
</s>