How to prove $\sinh^{-1} (\tan x)=\log \tan (\frac{\pi}{4}+\frac{x}{2})$ Like the question says
How to prove $$\sinh^{-1} (\tan x)=\log \tan (\frac{\pi}{4}+\frac{x}{2})$$
I have tried using many identity but in vain
For reference
$$\tanh ^{-1} x=\frac{1}{2} \log \frac{1+x}{1-x}$$
and $$\sinh^{-1} x=\log (x+\sqrt{x^2+1})$$

We note
\begin{align*}
\sinh^{-1} (\tan x)=\log \tan (\frac{\pi}{4}+\frac{x}{2})&\iff\tan x=\sinh\left(\log \tan (\frac{\pi}{4}+\frac{x}{2})\right).
\end{align*}
The latter equality follows because
\begin{align*}
\sinh\left(\log \tan (\frac{\pi}{4}+\frac{x}{2})\right)&=\frac{1}{2}\left(\tan (\frac{\pi}{4}+\frac{x}{2})-\cot (\frac{\pi}{4}+\frac{x}{2})\right)\\
&=\frac{1}{2}\left(\frac{\tan(x/2)+1}{1-\tan(x/2)}+\frac{\tan(x/2)-1}{1+\tan(x/2)}\right)\\
&=\frac{2\tan(x/2)}{1-\tan^2(x/2)}\\
&=\tan(x).
\end{align*}
