Derivation of $~\coth \left(\sinh^{-1} (x) \right)= {\sqrt{1+x^2} \over x }$

Question

I want to derive the following equation.

$$ \color{fuchsia}{\begin{align} \coth \left(\sinh^{-1} (\theta) \right)&= {\sqrt{1+\theta^2} \over \theta }~~~\text{for}~~\theta\in\mathbb{R}_{>0} \end{align}} $$

BTW since $~ \theta>0 ~$ is held, $~ \sinh^{-1}(\theta)>0 ~$ is also held.

I've prepared the following 2 eqns which can be used to derive the pink equation.

$$\begin{align} \coth(t)&= {{e^{t}+e^{-t}} \over e^{t}-e^{-t} }~~~ \text{for}~~t\in\mathbb{R}_{>0} \\ \sinh^{-1}(\theta)&=\ln \left(\theta+\sqrt{1+\theta^2} \right) \end{align}$$

$$\begin{align} y&:=\coth(t)\\ y&={e^{2t}+1 \over e^{2t}-1 }\\ \left(e^{2t}-1 \right)y&= e^{2t}+1\\ ye^{2t}-y&=e^{2t}+1\\ e^{2t}\left(y-1 \right)&=1+y\\ e^{2t}&= {y+1 \over y-1 }\\ y=1~~&\text{is not held under the condition of }~t>0\\ \ln \left(e^{2t} \right) &= \ln \left({y+1 \over y-1 } \right)\\ 2t&=\ln \left({\coth(t)+1 \over \coth(t)-1 } \right)\\ 2 \sinh^{-1}(\theta)&=\ln \left( { \coth(\sinh^{-1}(\theta))+1\over \coth(\sinh^{-1}(\theta))-1 } \right) \end{align}$$

I think my this approach can' be proceeded from here.

I have the hint that using $~ \exp \left(-\sinh^{-1}(t) \right)= {1 \over \theta+ \sqrt{1+\theta^{2}} } ~$ can(may) lead to derivation of the pink equation but I've had no idea how to utilize it and even can't derive that equation.

Which operation should I consider first?

Answer 1

It is much easier to think in terms of hyperbolic function identities rather than converting to exponentials and logs.

$$\begin{cases}\coth t = \frac{\cosh t}{\sinh t} \\ \cosh t = \sqrt{1+\sinh^2 t}\end{cases} \implies \coth(\sinh^{-1}(\theta)) = \frac{\sqrt{1+\sinh^2(\sinh^{-1}\theta)}}{\sinh(\sinh^{-1}\theta)} = \frac{\sqrt{1+\theta^2}}{\theta}$$

Answer 2

You have, $$e^t=\sqrt{1+\theta^2}+\theta\quad (1)$$ Take reciprocals, $$e^{-t}=\frac{1}{\sqrt{1+\theta^2}+\theta}$$ $$=\frac{1}{\sqrt{1+\theta^2}+\theta}\cdot\frac{\sqrt{1+\theta^2}-\theta}{\sqrt{1+\theta^2}-\theta}$$ $$=\sqrt{1+\theta^2}-\theta\quad(2)$$ From $(1)$ and $(2)$, you get $$e^t+e^{-t}=2\sqrt{1+\theta^2}$$ $$e^t-e^{-t}=2\theta$$ yielding the desired result. Hope this helps. :)