I want to prove the partial fraction decomposition:
\begin{align}\frac{\pi}{\sin \pi z} = \frac{1}{z} + 2z\sum\limits _{n=1} ^{\infty} \frac{(-1)^n}{z^2-n^2} \end{align}
with the help of the partial fraction decomposition of \begin{align} \pi \cot\pi z = \frac{1}{z} + 2z\sum\limits _{n=1} ^{\infty} \frac{1}{z^2-n^2}. \end{align}
Consider the Fourier series of $\cos{\alpha x}$, considered as a function on $[-\pi,\pi]$ (yes, this is perverse): it satisfies the Dirichlet conditions, so on the interior, we have $$ \cos{\alpha x} = \frac{1}{2}a_0+\sum_{k=1}^{\infty} a_k \cos{kx}, $$ where we find $a_k$ with the usual formula, $$ a_k = \frac{1}{\pi}\int_{-\pi}^{\pi} \cos{\alpha x} \cos{kx} \, dx = \frac{(-1)^k 2\alpha\sin{\pi\alpha}}{\pi(\alpha^2-k^2)}, $$ by an easy calculation. Oh dear, now it's fairly obvious what's about to happen: $$ \cos{\alpha x} = \frac{\sin{\pi\alpha}}{\pi \alpha}+\sum_{k=1}^{\infty} \frac{(-1)^k 2\alpha\sin{\pi\alpha}}{\pi(\alpha^2-k^2)} \cos{kx}. $$ Now set $x=0$ and multiply everything by $\pi/\sin{\pi \alpha}$. You can do the same thing for the cotangent by setting $x=\pi$.
To get from the cotangent to the cosecant, we go hunting for a trigonometric identity. Let's think about where the poles of each are: $\csc{\pi z}$ has poles at integers, as does $\cot{\pi z}$, but the residues alternate for the $\csc{\pi z}$. Therefore, can we make a linear combination of cotangents that has these poles? If you think for a few minutes, you come up with $$ \cot{\tfrac{1}{2}\pi z}-\cot{\pi z}, $$ which it is easy to verify works: $$ \frac{\cos{x}}{\sin{x}}-\frac{\cos{2x}}{\sin{2x}} = \frac{\cos{x}}{\sin{x}}-\frac{2\cos^2{x}-1}{2\sin{x}\cos{x}} = \frac{1}{2\sin{x}\cos{x}} = \csc{2x} $$ Now we just have to fiddle with the series expansions.
$$ \pi\cot{\tfrac{1}{2}\pi z}-\pi\cot{\pi z} = \frac{2}{z} + \sum_{n=1}^{\infty} \frac{z}{z^2/4-n^2}-\frac{1}{z} - \sum_{n=1}^{\infty} \frac{2z}{z^2-n^2} \\ =\frac{1}{z} + 2\sum_{n=1}^{\infty} \frac{2z}{z^2-(2n)^2} - \sum_{n=1}^{\infty} \frac{2z}{z^2-n^2}. $$ Ah, but hang on: the term with $n=2m$, even, in the second sum are half of the $m$th term of the first sum, whereas the odd terms have no corresponding terms. Hence the sums collapse into $$ \pi\csc{\pi z} = \frac{1}{z}+2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}. $$
Chappers' answer is very well-written and complete, but if you like another perspective, in order to prove that: $$ \frac{\pi}{\sin(\pi z)}=\frac{1}{z}+\sum_{n=1}^{+\infty}(-1)^n\left(\frac{1}{z-n}+\frac{1}{z+n}\right) $$ holds, we just need to prove that the residues of the LHS and of the RHS at the integers match.
Pretty easy: $$\text{Res}\left(\frac{\pi}{\sin(\pi z)},z=0\right) = \lim_{z\to 0}\frac{\pi z}{\sin(\pi z)}=1, $$
$$\text{Res}\left(\frac{\pi}{\sin(\pi z)},z=n\right) = \text{Res}\left(\frac{\pi}{\sin(\pi z+\pi n)},z=0\right)=(-1)^n\text{Res}\left(\frac{\pi}{\sin(\pi z)},z=0\right), $$ done.
Hint: Split both sums from your two formulas according to the parity of the iterator n, then notice that the former will be of the form $S_{\sin}=A-B$, and the latter $S_{\cot}=A+B$. Now, the value of B follows directly from your second formula, after a bit of factoring. Subtracting, we have A. Then replace both their values in the first, and voila ! :-$)$
