isolated singularity / Laurent series

Question

I want to classify the singularities of $f(z)=\frac{\cos^2 z}{\sin^2 z}$ Maybe I can write: $$f(z)=\frac{\cos^2 z}{\sin^2 z} = \frac{1-\sin^2 z}{\sin^2 z} = \frac{1}{\sin^2 z}-1.$$ I can substitute $\sin^2 z= \frac{1}{2}(1-\cos(2z))$. Then $\cos(2z)=1 $ for $ z=k\pi$ How can I go on from there?

Answer 1

For $z=kπ+z′$, where $z′≈0$, $sin^2(z)≈z′^2$, so pole of order $2$ for ALL $z=kπ$.