Singularities of $f(z) = \frac{z^2}{\cos(z) - 1}$

Question

I'm having difficulties approaching this function. If I am not mistaken $f(z) = \frac{z^2}{\cos(z) - 1}$ has singularities at $z = 2k\pi$. I can prove that the singularity $z = 0$ is removable but i can't figure out how to do it for $k \neq 0$. How can I prove it in order to find the Laurent series of $f(z)$ around its poles?