We have given,


As we know,
$\sinh ^{-1} x=\log x+\sqrt{1+x^2}$
So, $2 \sin \mathrm{h}^{-1} x=2 \log \left(x+\sqrt{1+x^2}\right)$
$\begin{aligned}
& =\log \left(x+\sqrt{1+x^2}\right)^2 \\
& =\log \left[x^2+1+x^2+2 x \sqrt{1+x^2}\right] \\
& 2 \sin \mathrm{h}^{-1} x=\log \left[2 x^2+1+2 x \sqrt{1+x^2}\right]
\end{aligned}$
Now, put $x=\frac{a}{\sqrt{1-a^2}}$
$\begin{aligned} & =\log \left[\frac{2 a^2}{1-a^2}+1+\frac{2 a}{\sqrt{1-a^2}} \times \frac{1}{\sqrt{1-a^2}}\right] \\ & =\log \left[\frac{2 a^2}{1-a^2}+1+\frac{2 a}{1-a^2}\right] \\ & =\log \left[\frac{2 a^2+1-a^2+2 a}{1-a^2}\right] \\ & =\log \left[\frac{a^2+2 a+1}{1-a^2}\right] \\ & \end{aligned}$
Now, by using Eq. (i),
$\begin{aligned}
& \log \left[\frac{a^2+2 a+1}{1-a^2}\right]=\log \left[\frac{1+x}{1-x}\right] \\
& \frac{a^2+2 a+1}{1-a^2}=\frac{1+x}{1-x}
\end{aligned}$
[by using componendo and dividendo rule]
$\begin{aligned}
& \frac{a^2+2 a+1+1-a^2}{a^2+2 a+1-1+a^2} \\
& =\frac{1+x+1-x}{1+x-1+x} \Rightarrow \frac{2+2 a}{2 a^2+2 a}=\frac{1}{x} \\
& \Rightarrow \frac{2(1+a)}{2 a(1+a)}=\frac{1}{x} \Rightarrow \frac{1}{a}=\frac{1}{x} \Rightarrow x=a
\end{aligned}$