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 $$\sinh^{-1} x=\log (x+\sqrt{x^2+1})$$
 A: Putting $\tan x$ in place of $x$ in this formula $$\sinh^{-1} x=\log (x+\sqrt{x^2+1})$$ we have, 
$$\sinh^{-1} \tan x=\log (\tan x+\sqrt{(\tan x)^2+1})$$
$$=\log (\tan x+\sqrt{(\sec x)^2})$$
$$=\log (\tan x+ \sec x)$$
$$=\log (\frac{1+\sin x}{\cos x})$$
$$=\log \left[\frac{(\sin \frac{x}{2}+\cos \frac{x}{2})^2}{\cos^2 \frac{x}{2}-\sin^2 \frac{x}{2}}\right]$$
$$=\log \left[\frac{\sin \frac{x}{2}+\cos \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right]$$
$$=\log \left[\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right]$$
$$=\log \left[\frac{\tan \frac{\pi}{4}+\tan \frac{x}{2}}{1-\tan \frac{\pi}{4}\tan \frac{x}{2}}\right]$$
$$=\log \tan (\frac{\pi}{4}+\frac{x}{2})$$
Hence proved.
A: Notice, $$\sinh^{-1}(\tan x)=\log(\tan x+\sqrt{\tan^2 x+1})$$
$$=\log(\tan x+\sec x)$$$$=\log\left(\tan x+\frac{1}{\cos x}\right)$$
$$=\log\left(\frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}+\frac{1+\tan^2\frac{x}{2}}{1-\tan^2\frac{x}{2}}\right)$$
$$=\log\left(\frac{\left(1+\tan\frac{x}{2}\right)^2}{\left(1-\tan\frac{x}{2}\right)\left(1+\tan\frac{x}{2}\right)}\right)$$
$$=\log\left(\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}\right)$$
$$=\log\left(\frac{\tan\frac{\pi}{4}+\tan\frac{x}{2}}{1-\tan\frac{\pi}{4}\tan\frac{x}{2}}\right)$$
$$=\color{red}{\log\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)}$$
A: 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*}
