I understand from this question that we technically cannot obtain a Laurent series for $f(z) = \frac{1}{\sin(z)}$, I just wanted to work through what I understand to be the steps to finding a Laurent series in general anyways before moving onto another method which better represents this function as a series.
Process below
Following the literature from the wiki page, I wanted to test it on a simple example, so I decided to try finding the expansion representation of $f(z) = \frac{1}{\sin(z)}$ where the Laurent series is defined as
$$\tag{1} p(z) =\sum_{n=-\infty}^\infty a_n(z-c)^n$$
where c is the center of the annulus and also the position of the singularity/singularities, and $a_n$ is defined as
$$\tag{2} a_n = \frac{1}{2\pi i} \oint \frac{f(z)}{(z-c)^{n+1}}dz$$.
Rewriting $\frac{1}{\sin(z)} = \frac{1}{(z-n\pi)}\frac{z-n\pi}{\sin(z)} = (z-n\pi)^{1+-1}\frac{1}{\sin(z)} = \frac{\frac{1}{\sin(z)}}{(z-n\pi)^{-1+1}}$, we plug this into (2), to get
$$\tag{3} a_{-1}=\frac{1}{2\pi i}\oint\frac{\frac{1}{\sin(z)}}{(z-n\pi)^{-1+1}}dz$$.
Rewriting again $\frac{1}{\sin(z)} = \frac{1}{(z-n\pi)}\frac{z-n\pi}{\sin(z)}$ and then expanding $\frac{z-n\pi}{\sin(z)}$ into a Taylor series representation centered around $n\pi$ yields: $$\frac{z-n\pi}{\sin(z)}=\frac{z-n\pi = 0 + 1(z-n\pi) + 0 + 0 + ...}{\sin(z) = 0+\cos(n\pi)(z-n\pi)-0-\frac{\cos(n\pi)(z-n\pi)^3}{6} + \frac{\cos(n\pi)(z-n\pi)^5}{120}+0+...} = \frac{z-n\pi}{(-1)^n\Big((z-n\pi)+\frac{1}{6}(z-n\pi)^3 + \frac{1}{120}(z-n\pi)^5+...\Big)}$$
Methods for dividing these 2 series varies, but the results are the same. I went with long division which yielded:
$$\frac{z-n\pi}{\sin(z)}= (-1)^n\Big(1(z-n\pi)^0+\frac{1}{6}(z-n\pi)^2 + \frac{7}{360}(z-n\pi)^4+...\Big)$$
And then we just multiply this result by $\frac{1}{z-n\pi}$ to get the expansion for $\frac{1}{\sin(z)}$
And plugging this result back into (3) gives
$$\tag{4} a_{n} = \frac{1}{2\pi i}\oint \frac{(-1)^n\Big(\frac{1}{(z-n\pi)}+\frac{1}{6}(z-n\pi)^1 + \frac{7}{360}(z-n\pi)^3+...\Big)}{(z-n\pi)^{-1+1}} dz$$.
If I understand this correctly, using the Cauchy Residue theorem lets us evaluate this integral by simply finding the coefficient attached to $(z-c)^n$ inside of the Taylor series such that n is the same as n in the denominator in (3). In this case, it would be the coefficient of our first term, $(-1)^n$, $(z-n\pi)^{-1}$ is raised to the same exponent. This means (3) evaluates to:
$$a_{-1} = (-1)^n$$. And plugging this result into (1) and rewriting it a bit, we get:
$$\tag{5} p(z) = \sum_{n=-\infty}^\infty (-1)^n(z-n\pi)^{-1} \\ = \sum_{n=-\infty}^\infty \frac{(-1)^n}{(z-n\pi)} \\ = \frac{1}{z} + \sum_{k=1}^\infty (-1)^n\Big(\frac{1}{z+n\pi}+\frac{1}{z-n\pi}\Big) \\ = \frac{1}{z} + \sum_{k=1}^\infty (-1)^n\Big(\frac{z-n\pi}{(z+n\pi)(z-n\pi)}+\frac{z+n\pi}{(z-n\pi)(z+n\pi)}\Big) \\ = \boxed{\frac{1}{z} + 2z\sum_{k=1}^\infty (-1)^n \frac{1}{z^2+(n\pi)^2}}$$
NOTE: I notice my result matches the pole expansion $\csc(z)$ which can be found here... I found the Laurent series, right? Or did I find the Pole Expansion, or are they the same thing? Would like to also get my definitions straight before moving on.