How to prove the equality $\arcsin(\tanh x) = \arctan(\sinh x)$?

Question

I understand that these are equal, but don't know how to go about proving it. I only got as far as replacing the hyperbolic functions with $e$, not sure what to do afterwards.

Answer 1

Let $y= \sin^{-1}(\tanh(x))$ so $\sin y =\tanh (x)$, square this and subtract it from $1$ \begin{eqnarray*} \cos^2(y)=1-\sin^2(y)=1-\tanh^2(x)=\operatorname{sech}^2(x) \end{eqnarray*} Now recipricate this and subtract $1$ \begin{eqnarray*} \tan^2(y)=\sec^2-1=\cosh^2(x)-1=\sinh^2(x) \end{eqnarray*} Now square root this and take the inverse tan and we have $y= \tan^{-1}(\sinh(x))$, thus \begin{eqnarray*} \color{blue}{\sin^{-1}(\tanh(x))=\tan^{-1}(\sinh(x))}. \end{eqnarray*}

Answer 2

Hint

$$\arcsin\left(\frac{x}{\sqrt{1+x^2}}\right)=\arctan\left(x\right)$$

Answer 3

$\DeclareMathOperator{\th}{th}\DeclareMathOperator{\sh}{sh}\DeclareMathOperator{\ch}{ch}$

Let's set $\begin{cases} a=\arcsin(\th(x))\in[-\frac{\pi}2,\frac{\pi}2]\implies\th(x)=\sin(a)\\ b=\arctan(\sh(x))\in[-\frac{\pi}2,\frac{\pi}2]\implies\sh(x)=\tan(b)\\ \end{cases}$

First note that $\th,\sh,\sin\tan$ are all odd functions.

So $\th(x)$ and $\sh(x)$ have the same sign, thus given the range for $a$ and $b$, they also have the same sign.

$\cos(a)^2=1-\sin(a)^2=1-\th(x)^2=\dfrac 1{\ch(x)^2}=\dfrac 1{1+\sh(x)^2}=\dfrac 1{1+\tan(b)^2}=\cos(b)^2$

Given the range of $a$ and $b$ then $\cos(a)\ge 0$ and $\cos(b)\ge 0$

So $\cos(a)=\cos(b)$ and since $a$ and $b$ have the same sign, then $a=b$. $\square$

Answer 4

Both the LHS and the RHS are odd functions in $C^{\infty}(\mathbb{R})$ and every $x\in\mathbb{R}$ can be written as $\log u$ for some $u\in\mathbb{R}^+$. It follows that it is enough to check $$\forall u\in\mathbb{R}^+,\qquad \arcsin\tanh\log u = \arctan\sinh\log u $$ or $$ \forall u\in\mathbb{R}^+,\qquad \arcsin \frac{u^2-1}{u^2+1} = \arctan \frac{u^2-1}{2u}.$$ Both these functions belong to $C^\infty(\mathbb{R}^+)$ and fulfill $\lim_{u\to 0^+}f(u)=-\frac{\pi}{2}$.
Moreover, they both fulfill $$ f\left(\tfrac{1}{u}\right) = -f(u) $$ hence it is enough to check the previous identity over $u\geq 1$.
By letting $u=v+\sqrt{v^2+1}$ the problem boils down to showing $$ \forall v\geq 0, \qquad \arcsin\frac{v}{\sqrt{1+v^2}} = \arctan v $$ which is pretty trivial from the following diagram: