How to prove $\sinh^{-1} (\tan x)=\log \tan (\frac{\pi}{4}+\frac{x}{2})$

Question

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

Answer 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.

Answer 2

Notice, $$\sinh^{-1}(\tan x)=\log(\tan x+\sqrt{\tan^2 x+1})$$ $$=\log(\tan x+\sec x)$$$$=\log\left(\tan x+\frac{1}{\cos x}\right)$$ $$=\log\left(\frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}+\frac{1+\tan^2\frac{x}{2}}{1-\tan^2\frac{x}{2}}\right)$$ $$=\log\left(\frac{\left(1+\tan\frac{x}{2}\right)^2}{\left(1-\tan\frac{x}{2}\right)\left(1+\tan\frac{x}{2}\right)}\right)$$ $$=\log\left(\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}\right)$$ $$=\log\left(\frac{\tan\frac{\pi}{4}+\tan\frac{x}{2}}{1-\tan\frac{\pi}{4}\tan\frac{x}{2}}\right)$$ $$=\color{red}{\log\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)}$$

Answer 3

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*}