If 2 ^ -1 ( a 1-a^2 )= ((1+x)/(1-x)), then x is equal to

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