Simplifying $\sinh\left(2\operatorname{arctanh}\left(e^{Hx}\right)\right)$ Is there anyone who can simplify this expression 
$$\sinh\left(2\operatorname{arctanh}\left(e^{Hx}\right)\right)$$ 
Is it possible to this equal to be $-\operatorname{csch}(Hx)$ ? 
In Wikipedia's "Hyperbolic functions" entry I saw that
$$\operatorname{arctanh}(x) = \frac{1}{2}\ln\frac{1+x}{1-x}$$
I tried to simplify it but I could not ...
 A: Yes, it's possible to be equal. Using what you found and the hyperbolic function definitions, you have
$$\begin{equation}\begin{aligned}
\sinh\left(2\operatorname{arctanh}\left(e^{Hx}\right)\right) & = \sinh\left(\ln\frac{1+e^{Hx}}{1-e^{Hx}}\right) \\
& = \frac{e^{\ln\frac{1+e^{Hx}}{1-e^{Hx}}} - e^{-\ln\frac{1+e^{Hx}}{1-e^{Hx}}}}{2} \\
& = \frac{\frac{1+e^{Hx}}{1-e^{Hx}} - \frac{1-e^{Hx}}{1+e^{Hx}}}{2} \\
& = \frac{\frac{1+2e^{Hx}+e^{2Hx}-(1-2e^{Hx} + e^{2Hx})}{(1-e^{Hx})(1+e^{Hx})}}{2} \\
& = \frac{\frac{4e^{Hx}}{1-e^{2Hx}}}{2} \\
& = \frac{2e^{Hx}}{{1-e^{2Hx}}}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Also, you have
$$\begin{equation}\begin{aligned}
-\operatorname{csch}(Hx) & = -\frac{2e^{Hx}}{e^{2Hx} - 1} \\
& = \frac{2e^{Hx}}{1 - e^{2Hx}}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
As you can see, \eqref{eq1A} and \eqref{eq2A} match.
A: If $t=\tanh\frac{u}{2}$ with $u:=2\operatorname{arctanh}e^{Hx}\implies t=e^{Hx}$, we can use the double-hyperbolic-angle formula $\sinh u=\frac{2t}{1-t^2}=\frac{-2}{e^{Hx}-e^{-Hx}}=-\operatorname{csch}Hx$.
A: Let $u=\tanh t$. Then
$$
u=\frac{e^t-e^{-t}}{e^t+e^{-t}}=\frac{e^{2t}-1}{e^{2t}+1}=
1-\frac{2}{e^{2t}+1}
$$
so
$$
e^{2t}+1=\frac{2}{1-u}
$$
and therefore
$$
t=\operatorname{artanh}u=\frac{1}{2}\log\frac{1+u}{1-u}=
\frac{1}{2}\log f(u)
$$
In particular, with $u=e^{Hx}$,
\begin{align}
\sinh(2\operatorname{artanh}(e^{Hx})
&=\sinh(2\operatorname{artanh}(u))\\[2ex]
&=\sinh(\log f(u))\\[2ex]
&=\frac{1}{2}\left(f(u)-\frac{1}{f(u)}\right)\\
&=\frac{1}{2}\left(\frac{1+u}{1-u}-\frac{1-u}{1+u}\right)\\
&=-\frac{2u}{u^2-1}\\
&=-\frac{2}{u-u^{-1}}\\
&=-\frac{2}{e^{Hx}-e^{-Hx}}\\
&=-\frac{1}{\sinh(Hx)}
\end{align}
A: Using the tangent half-angle substitution $$\sinh \left(2 \tanh ^{-1}(t)\right)=\frac{2 t}{1-t^2}$$
$$t=e^{Hx} \implies \sinh\left(2\tanh ^{-1}\left(e^{Hx}\right)\right)=-\frac{2 e^{H x}}{e^{2 H x}-1}=-\frac{2 }{e^{ H x}-e^{-H x}}$$ that is to say
$$-\frac 1{\sinh(Hx)}$$
