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 $$\sinh^{-1} x=\log (x+\sqrt{x^2+1})$$

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.
