Prove csc⁡(x)=∑k=−∞∞(−1)kx+kπ

We can use also this well known summation formula (a consequence of the residue theorem):
${\displaystyle \sum _{k\in \mathbb{Z}}{(-1)}^{k}f\left(k\right)=-\sum \left[\text{residues of }\pi \csc \left(\pi z\right)f\left(z\right)at{f\left(z\right)}^{\prime }s\text{ poles}\right]}$
so if we take ${\displaystyle f\left(z\right)=\frac{1}{x+z\pi }}$ we get
${\displaystyle \sum _{k\in \mathbb{z}}\frac{{(-1)}^k}{x+k\pi }=-Re{s}_{z=-\frac{x}{\pi }}\left(\frac{\pi \csc \left(\pi z\right)}{x+z\pi }\right)=\csc \left(x\right)}$
as wanted.

In this answer (using complex methods) and in this answer (using real methods), it is shown in detail that
${\displaystyle \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{z+k}=\pi \cot \left(\pi z\right)}$ (1)
(1) is the sum for even and odd k. The sum for even k would be
${\displaystyle \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{z+2k}=\frac{\pi }{2}\cot \left(\frac{\pi z}{2}\right)}$ (2)
The sum for even minus the sum for odd would be twice (2) minus (1)
${\displaystyle \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{(-1)}^k}{z+k}=\pi \cot \left(\frac{\pi z}{2}\right)-\pi \cot \left(\pi z\right)}$
${\displaystyle =\pi \frac{1+\cos \left(\pi z\right)}{\sin \left(\pi z\right)}-\pi \frac{\cos \left(\pi z\right)}{\sin \left(\pi z\right)}}$
${\displaystyle =\pi \csc \left(\pi z\right)}$
Therefore,
${\displaystyle \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{(-1)}^k}{z+k\pi }=\csc \left(z\right)}$

We begin by expanding the function $\cos (xy)$ in a Fourier series, $\cos (xy)=a_0/2+\sum _{k=1}^{\mathrm{\infty }}{a}_k\cos (ky)$ (1) for $x\in [-\pi /\pi ]$. The Fourier coefficients (1) are given by $a_k=\frac{2}{\pi }{\int }_0^{\pi }\cos (xy)\cos (ky)dy$ $=\frac{1}{\pi }(-1{)}^k\sin (\pi x)(\frac{1}{x+k}+\frac{1}{x-k})$ (2) Substituting (2) into (1), setting $y=0$, and dividing by $\sin (\pi x)$ reveals $\pi \csc (\pi x)=\frac{1}{y}+\sum _{n=1}^{\mathrm{\infty }}(-1{)}^k(\frac{1}{x-k}+\frac{1}{x+k})$ $=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{(-1{)}^k}{x-k}$ $=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{(-1{)}^k}{x+k}$ (3) Finally, enforcing the substitution $x\to x/\pi$ and dividing by $\pi$ in (3) yields the coveted result $\csc (x)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{(-1{)}^k}{x+k\pi }$