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:
$$\tan^{-1} \left(\dfrac {\sqrt {1+x^2}-\sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) =\alpha$$
$$\dfrac {\sqrt {1+x^2}-\sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}=\tan (\alpha )$$
$$\dfrac {1+x^2-2\sqrt {1+x^2}.\sqrt {1-x^2}+ 1 - x^2}{1+x^2-1+x^2}=\tan (\alpha)$$
$$\dfrac {1-\sqrt {1+x^2}.\sqrt {1-x^2}}{x^2}=\tan (\alpha)$$
 A: Continuing from where you stopped 
We get 
$$1-x^2\tan\alpha= \sqrt {1+x^2} .\sqrt {1-x^2}$$
Squaring both sides we get 
$$1+x^4\tan^2\alpha-2x^2\tan\alpha=1-x^4$$
$$x^4(1+\tan^2\alpha)=2x^2\tan\alpha$$
$$x^2=\frac {2\tan\alpha}{1+\tan^2\alpha}$$
Hence $$x^2=\sin(2\alpha)$$
A: 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}
A: 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}
A: 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
$x^2=\cos2y=\cos\left(2m\pi+\dfrac\pi2-2\alpha\right)=?$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}
