I evaluated the Fourier series for $e^{-ax}$ , where $ -\pi \le x \le \pi$, which comes out be what is the first equation in picture tagged with this post below now what I am expected to do is to find the equation for:
$${\cfrac{\pi}{sinh\;\pi}}$$
to do that it is made to deduct using $a = 1$ and $x = 0$, my question is why did not we use other combinations like:
$a = 0, x= 0$
$ a = 0, x=1$
as these combinations also make
$e^{-ax}=1.$
As your problem says $$e^{-ax} = \frac{2\sinh(a\pi)}{\pi}\sum_n\left(c_n\cos(nx) - d_n\sin(nx)\right)$$ Now to get $$\frac{\pi}{\sinh(\pi)}$$ you have to choose the right coefficients: first $a$ has to be $1$ so that $$\sinh(a\pi) = \sinh(\pi)$$ also it cannot be zero due to the fact that the first element of the cosine expansion is $\frac{1}{2a}$ that is not defined in zero.
Choosing $x=0$ is just so that the $\sin(nx)$ expansion is cancelled and $e^{-ax}=1$, simplifying the expression even more.