Sum involving $\cosh$ and $\sinh$

I would like to prove the equation

$$\frac{\sinh\left(\left (1-\frac{1}{2m} \right)x\right)}{\sinh(x/2m)}=1+ \sum\limits_{n=1}^{m-1}2\cdot \cosh\left(\left( 1-n/m \right)x\right),\quad \forall x > 0, m\in\mathbb{N}.$$

where the sum is evaluated as $0$ for $m=1$.

But I do not see, how to approach this problem. Proof by induction seems not to be appropriate, and it does not seem to be a simple application of the standard addition formulas. Does anyone has an idea?

Best wishes

Substitute $x$ by $mx$. Write $\frac{1}{2}(e^x-e^{-x})$ instead of $\sinh(x)$ and $\frac{1}{2}(e^x+e^{-x})$ instead of $\cosh(x)$. Then you can calculate the right sum by using $\sum\limits_{n=1}^{m-1}z^n=z\frac{z^{m-1}-1}{z-1}$. At the end substitute $x$ by $\frac{x}{m}$.
We have $$\sum_{n=1}^{m-1}2\cosh\left(\left(1-\frac{n}{m}\right)x\right)+1=e^{x}\sum_{n=1}^{m-1}\left(e^{-x/m}\right)^{n}+e^{-x}\sum_{n=1}^{m-1}\left(e^{x/m}\right)^{n}+1$$ $$=\frac{\left(e^{x}-e^{x/m}\right)\left(1+e^{-x}\right)}{e^{x/m}-1}+1=\color{red}{\frac{\sinh\left(x-\frac{x}{2m}\right)}{\sinh\left(\frac{x}{2m}\right)}}$$ as wanted.