How do I find the Laurent series for $\frac{1}{(1+\cos z)^2}$ around $z_0=\pi$?

Question

What I tried so far:
Using that $\cos z = \sum_{n=0}^{\infty}\dfrac{(-1)^n}{(2n)!}z^{2n}$ , for any $z \in \mathbb C$, then, for $0<|z-\pi|<1$: $$\begin{aligned}\dfrac{1}{(1+\cos z)^2} &= \dfrac{1}{(1+\cos ((z-\pi)+\pi))^2} \\&= \dfrac{1}{(1-\cos(z-\pi))^2}\\ &=\dfrac{1}{\left(1-\left(1-\dfrac{(z-\pi)^2}{2} + \dfrac{(z-\pi)^4}{24}-\dfrac{(z-\pi)^6}{720}+...\right)\right)^2}\\ &=\dfrac{1}{\left(\dfrac{(z-\pi)^2}{2} - \dfrac{(z-\pi)^4}{24}+\dfrac{(z-\pi)^6}{720}-...\right)^2} \end{aligned}$$ Now I don't know how to proceed (and am I even doing things right? I don't know if this is a rigorous development).

Answer 1

May be useful the identity $$\frac{1}{(1+\cos z)^2}=\frac{1}{4} \sec ^4\left(\frac{z}{2}\right)$$ and at $z=\pi$ $$\sec \left(\frac{z}{2}\right)=-\frac{2}{z-\pi }-\frac{z-\pi }{12}-\frac{7 (z-\pi )^3}{2880}+O\left(z^5\right)$$

Answer 2

If, as you wrote in the comments, you're after a proof of the fact $\operatorname{res}_{z=\pi}\frac1{(1+\cos z)^2}$, using the Laurent series, note that if$$\frac1{(1+\cos z)^2}=\sum_{n=-\infty}^\infty a_n(z-\pi)^n,$$then\begin{align}\sum_{n=-\infty}^\infty(-1)^na_n(z-\pi)^n&=\sum_{n=-\infty}^\infty a_n\bigl((2\pi-z)-\pi\bigr)^n\\&=\frac1{(1+\cos(2\pi-z))^2}\\&=\frac1{(1+\cos z)^2}\\&=\sum_{n=-\infty}^\infty a_n(z-\pi)^n\end{align}and therefore, if $n$ is odd, then $a_n=0$. In particular, $a_{-1}=0$.