How to prove $\tanh ^{-1} (\sin \theta)=\cosh^{-1} (\sec \theta)$ As the question says 
How to prove 
$$\tanh ^{-1} (\sin \theta)=\cosh^{-1} (\sec \theta)$$
I have tried to solve it
The end result that got for RHS
$$=\log \frac{1+\tan\frac{\theta}{2}}{1-\tan \frac{\theta}{2}}$$
I am stuck here Please help

We know by formula,$$\tanh^{−1}x=\frac{1}{2}\log\left(\frac{1+x}{1-x}\right)$$ and $$\cosh^{-1} x=\log (x+\sqrt{x^2-1})$$
Now putting $x=\sin \theta$ in the formula for $\tanh ^{-1}x$, we have that 
$$\tanh^{−1}(\sin \theta)=\frac{1}{2}\log\left(\frac{1+\sin \theta}{1-\sin \theta}\right)$$
$$=\frac{1}{2}\log\left(\frac{(\cos \frac{\theta}{2}+\sin \frac{\theta}{2})^2}{(\cos \frac{\theta}{2}-\sin \frac{\theta}{2})^2}\right)$$
$$=\frac{1}{2}\log\left(\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}\right)^2$$
$$=\frac{1}{2}\cdot 2\log\left(\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}\right)$$
$$=\log\left(\frac{(\cos \frac{\theta}{2}+\sin \frac{\theta}{2})}{(\cos \frac{\theta}{2}-\sin \frac{\theta}{2})}\cdot \frac{(\cos \frac{\theta}{2}+\sin \frac{\theta}{2})}{(\cos \frac{\theta}{2}+\sin \frac{\theta}{2})}\right)$$
$$=\log\left(\frac{(\cos \frac{\theta}{2}+\sin \frac{\theta}{2})^2}{\cos^2 \frac{\theta}{2}-\sin^2 \frac{\theta}{2}}\right)$$
$$=\log (\frac{1+\sin \theta}{\cos \theta})$$
$$=\log (\tan \theta+ \sec \theta)$$
$$=\log (\sec \theta+\sqrt{(\tan \theta)^2})$$
$$=\log (\sec \theta+\sqrt{(\sec \theta)^2-1})$$
$$=\cosh^{-1} (\sec \theta)$$
Hence proved, assuming $0 < \theta < \frac{\pi}{2}$
EDIT: The proof is valid for $0 < \theta < \frac{\pi}{2}$ but not in general.
