Laurent expansion of $1/\sin^2(z)$.

I am trying to compute the Laurent expression of the function $f(z) = 1/\sin^2(z)$, and I am unable to see where I went wrong. This is my work so far:

By checking WolframAlpha, one sees that the Laurent series does have a principal part (with terms $a_n z^{n}, n<0$), and therefore I expect to get a Laurent series with some negative powers. I am interested in the annulus defined by $\pi < |z| < 2\pi$, and we will need to keep this in mind because trigonometric functions are not bound in the complex plane. What I did is this:

$$\frac{1}{\sin^2(z)} = \frac{1}{1-\cos^2(z)} = (…)$$

Because $|\cos^2(z)|>1$ when $\pi < |z| < 2\pi$, we will need to manipulate the expression a little. Multiplying and dividing the denominator by $\cos^2(z)$, we get: $$(…) = -\frac{1}{\cos^2(z)}\frac{1}{1-\frac{1}{\cos^2(z)}} = (…)$$ I was able to calculate the first few terms of $1/\cos(z)$, and as I result I know that $1/\cos^2(z) = 1 + z^2 + \frac{2}{3}z^4 + O(z^6)$. Since we know that for $|w|<1$ $$\frac{1}{1-w} = \sum_{n=0}^\infty w^n$$ Then I arrive at:

$$(…) = -(1+z^2+O(z^4))\cdot (1 + 1+z^2+O(z^4))$$

It is clear that this product will not yield negative powers anywhere. Where did I go wrong? I suspect it is in my assumption about the values that the cosine will take in the annulus, since in the vicinity of the real axis it does take small values. I do not know how to arrive to the correct Laurent series. I have been trying to solve this in many different ways, but I do not get the desired result. Any help will be appreciated.

Given a pole at $z_0$, you can set $w=z-z_0$ so $z=w+z_0$, then letting $g(w)=f(w+z_0)$ you can expand around $w=0$ to make things easier to notate.

Noting first that $\sin^2(z)= \frac{1-\cos(2z)}{2}$, which gives an easy well known series to work with when expanding.

We have that $\sin^2(n\pi)=0$ for each $n \in \mathbb{Z}$, so $f$ has a pole at each $z_n=n\pi$. Then letting $w=z-z_n$, we have around $0$ that:

$$\begin{align} g(w) & = f(z+n\pi) \\ & = 1/\sin^2(z+n\pi) \\ & = \dfrac{2}{1-\cos(2z+2n\pi)} \\ &= \dfrac{2}{1-\cos(2z)} \\ & = \dfrac{2}{1-1+\dfrac{(2z)^2}{2}-\dfrac{(2z)^4}{24}+O(z^6)}\\ & = z^{-2}*\dfrac{1}{1-\dfrac{z^2}{3}+O(z^4)}\\ & = z^{-2}*\sum_{n=0}^\infty(\dfrac{z^2}{3}+O(z^4))^n \\ & = z^{-2}*(1+z^2/3+O(z^4)) \\ & = \dfrac{1}{z^2}+\dfrac{1}{3}+O(z^2) \end{align}$$

This'll work at least near $n\pi$, but again we'll have issues on whether $\cos(2z)$ will be sufficiently bounded.