# 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

• Please check my solution to your previous problem on hyperbolic functions and also my solution to this one. Read them carefully and I am sure you will then be able to carry out any other problem you might face in this topic yourself. If you have any doubts, please clarify them for your own good. – SchrodingersCat Nov 30 '15 at 7:28
• I would be a bit suspicious about this one. The left-hand side is defined and real for all real $\theta$. The right-hand side is not. Arcosh is real only for (real) arguments greater than 1. The Secant function is negative in a lot of places. – mickep Nov 30 '15 at 7:36
• i have not thought about any of that, i just saw question in a book and trying to solve. @mickep – Vinay5forPrime Nov 30 '15 at 7:41
• A good exercise is then: Try to find out why the seemingly correct calculation of @Aniket is not true in general. – mickep Nov 30 '15 at 7:46
• @mickep is correct. My calculation was based on the assumption that $0 < \theta < \frac{\pi}{2}$. Forgot to mention it earlier but edited it now. – SchrodingersCat Nov 30 '15 at 7:52

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.

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.

• I hope you do not mind that I added a graph visualizing your argument. – mickep Nov 30 '15 at 7:40
• @RobertIsrael Possible to see an identity (in place of a verbal description) with an added term of angle rotation $\pi n ?$ . – Narasimham Nov 30 '15 at 12:28

Notice, the equality is true for $0\le \theta<\pi/2$ $$LHS=\tanh^{-1}(\sin\theta)$$$$=\frac{1}{2}\log\left(\frac{1+\sin\theta}{1-\sin\theta}\right)$$ $$=\frac{1}{2}\log\left(\frac{\sec\theta(1+\sin\theta)}{\sec\theta(1-\sin\theta)}\right)$$ $$=\frac{1}{2}\log\left(\frac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta}\right)$$ $$=\frac{1}{2}\log\left(\frac{(\sec\theta+\tan\theta)(\sec\theta+\tan\theta)}{(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)}\right)$$ $$=\frac{1}{2}\log\left(\frac{(\sec\theta+\tan\theta)^2}{\sec^2\theta-\tan^2\theta}\right)$$ $$=\frac{1}{2}\log\left(\frac{(\sec\theta+\tan\theta)^2}{1}\right)$$ $$=\frac{2}{2}\log\left(\sec\theta+\tan\theta\right)$$ $$=\log\left(\sec\theta+\sqrt{\sec^2\theta-1}\right)$$ $$=\cosh^{-1}(\sec\theta)=RHS$$

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.