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
 A: It's true for $0 \le \theta < \pi/2$, but not in general.  For $\pi/2 < \theta < 3 \pi/2$, $\sec(\theta) < 0$ so $\cosh^{-1}(\sec(\theta))$ is not real, although $\tan^{-1}(\sin(\theta))$ is.  For $3 \pi/2 < \theta < 2 \pi$, $\sin(\theta) < 0$ so $\tanh^{-1}(\sin(\theta)) < 0$, while $\sec(\theta) > 1$ and $\cosh^{-1}(\sec(\theta)) > 0$.  In fact, $\tanh^{-1}(\sin(\theta))$ is an odd function, while $\cosh^{-1}(\sec(\theta))$ is an even function.  
Edit by mickep:
Here is a plot of both $\text{artanh}\,\sin\theta$ (blue) and $\text{arcosh}\,\sec\theta$ (yellow) for $-10<\theta<10$. I guess it can help you to visualize the argument of Robert.

A: 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.
A: Notice, the equality is true for $0\le \theta<\pi/2$ $$LHS=\tanh^{-1}(\sin\theta)$$$$=\frac{1}{2}\log\left(\frac{1+\sin\theta}{1-\sin\theta}\right)$$
$$=\frac{1}{2}\log\left(\frac{\sec\theta(1+\sin\theta)}{\sec\theta(1-\sin\theta)}\right)$$
$$=\frac{1}{2}\log\left(\frac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta}\right)$$
$$=\frac{1}{2}\log\left(\frac{(\sec\theta+\tan\theta)(\sec\theta+\tan\theta)}{(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)}\right)$$
$$=\frac{1}{2}\log\left(\frac{(\sec\theta+\tan\theta)^2}{\sec^2\theta-\tan^2\theta}\right)$$
$$=\frac{1}{2}\log\left(\frac{(\sec\theta+\tan\theta)^2}{1}\right)$$
$$=\frac{2}{2}\log\left(\sec\theta+\tan\theta\right)$$
$$=\log\left(\sec\theta+\sqrt{\sec^2\theta-1}\right)$$
$$=\cosh^{-1}(\sec\theta)=RHS$$
A: Let 
$$\tanh ^{-1} (\sin \theta)= \cosh^{-1} (\sec \theta) = x $$
gives
$$\tanh  x = \sin \theta ;\,  \cosh x = \sec \theta ; \; \sech x = \cos\theta ;\;$$
Identities that hold good for all $x,\theta $ 
$$ \sin^2..+ \cos^2.. = 1 ; \; \sech^2 .. + \tanh^2 .. =1 ; $$
This would make us temporarily believe that the given equation is an identity.
An examination of the given functions however shows identity validity for certain intervals only as stated by Robert Israel and graphed by mickup. This restriction is due to inverse functions and square roots not taking negative arguments.The arccosh function is even, everywhere positive with sudden slope discontinuities and  a restricted real existence within odd multiples of $\pi/2$ arguments including co-terminal periodicity.
