Consider the function\begin{eqnarray*}f_{a}(z)=\left(1-a\right)\cdot \frac{z\cdot \left(\tanh\left(a\cdot z\right)+\tanh\left(z\right)\right)}{\ln\left(\cosh\left(a\cdot z\right)\right)-\ln\left(\cosh\left(z\right)\right)}\end{eqnarray*}for $z\geq 0$ and $a\ne 1$. It is not so difficult to show that all values of $f_{a}(z)$ are negative for all $z\geq 0$ and all admissible $a$. Moreover $\lim_{z\to 0}f_{a}(z)=-2$.
Maple shows that for $a\leq 0$ the function $f$ is strictly increasing, with $\lim_{z\to\infty}f_{a}(z)=0$, while for $a>0$ we obtain $\lim_{z\to\infty}f_{a}(z)=-2$ with a unique maximum and no other critical points.
The Maple result seems to be difficult to prove. I would be very grateful for any hint.
Note that
\begin{eqnarray*}\frac{d}{dt}\ln\left(\cosh\left(t z\right)\right)=z\tanh(t z).\end{eqnarray*}
