how to prove $\left( {\frac{1}{\sin z}} \right)^{2}=\sum_{-\infty}^{+\infty}\frac{1}{\left( {z-\pi n} \right)^2}$
Prove that $\frac{1}{\sin^2 z } = \sum\limits_{n= -\infty} ^ {+\infty} \frac{1}{(z-\pi n)^2} $