I want to prove:
$$\frac{1}{\sin z} = \frac{1}{z} + \sum_{n=1}^{\infty} \frac{(-1)^n 2z}{z^2 - \pi^2 n^2}$$
I tried to prove, I got this: $g(z) := \frac{1}{z} + \sum_{n=1}^{\infty} \frac{(-1)^n 2z}{z^2 - \pi^2 n^2} = \sum_{- \infty}^{+ \infty} \frac{(-1)^n}{z+n\pi}$ so $g(z)$ and $h(z) := \frac{1}{\sin z}$ have same pole of same order so $k(z) = h(z)-g(z)$ is holomorphic map. but I can't complete this.
$$k(z)=\frac{1}{\sin z} -\frac{1}{z} - \sum_{n=1}^{\infty} \frac{(-1)^n 2z}{z^2 - \pi^2 n^2}=\frac{1}{\sin z}-\sum_{n=- \infty}^{+ \infty} \frac{(-1)^n}{z+n\pi}$$ is meromorphic, it has no pole so it is entire, it is $2\pi$-periodic, since $k(z)$ is odd then $k(0)=0$ and $$w(z)=\frac{k(z)}{z}=\frac{1}{z\sin z} -\frac{1}{z^2} - \sum_{n=1}^{\infty} \frac{(-1)^n 2}{z^2 - \pi^2 n^2}$$ is entire, by absolute convergence it is bounded on $\Re(z)\in[-\pi,\pi]$, thus $k(z)=O(z)$ on $\Re(z)\in[-\pi,\pi]$, thus $k(z)=O(z)$ on the whole complex plane, thus $w(z)$ is a bounded entire function, thus it is constant $$w(z)=w(2\pi)=\frac{k(2\pi)}{2\pi }=\frac{ k(0)}{2\pi }=0 $$