Expansion of $\frac{1}{ \sin x} $ How is
$$\csc x = \dfrac{1}{t} + \sum\limits_{k= 1}^{+ \infty} (-1)^k \Big( \dfrac{1}{t + kπ} + \dfrac{1}{t - kπ} \Big) $$
Where $x = k \pi ± t$
I saw this in one of the proofs and I'm not sure how? It says it is the partial fraction form of $\csc x $ 

We begin by expanding the function $\cos(ax)$ in a Fourier series, 
$$\cos(xy)=a_0/2+\sum_{n=1}^\infty a_n\cos(nx) \tag1$$
for $x\in [-\pi/\pi]$.  The Fourier coefficients are given by
$$\begin{align}
a_n&=\frac{2}{\pi}\int_0^\pi \cos(xy)\cos(nx)\,dx\\\\
&=\frac1\pi (-1)^n \sin(\pi y)\left(\frac{1}{y +n}+\frac{1}{y -n}\right)\tag 2
\end{align}$$
Substituting $(2)$ into $(1)$, setting $x=0$, and dividing by $\sin(\pi y)$ reveals
$$\begin{align}
\pi \csc(\pi y)&=\frac1y +\sum_{n=1}^\infty (-1)^n\left(\frac{1}{y -n}+\frac{1}{y +n}\right) \tag 3
\end{align}$$
Dividing by $\pi$ and setting $y=t/\pi$  yields
$$\csc(t)=\frac1{t} +\sum_{n=1}^\infty (-1)^n\left(\frac{1}{t -n\pi}+\frac{1}{t +n\pi }\right)$$
