Type of Singularity for $ \dfrac{\cos z}{(1-\sin z)} $ at $z=\pi /2$.

Question

What is the type of singularity for the following function:

$ \dfrac{\cos z}{(1-\sin z)} $ at $z=\pi /2 $ ?

I substituted $z = t+\pi / 2$ and then used expansions for $\cos z$ and $\sin z$ respectively. I was not able to reduce it to the form as Laurent series. I think different approach should be used to solve it.

Answer 1

By substituting $z\mapsto \frac{\pi}{2}-w$, the question boils down to finding the type of singularity of $$ \frac{\sin w}{1-\cos w} = \frac{\cos\tfrac{w}{2}}{\sin\tfrac{w}{2}}=\cot\tfrac{w}{2} $$ at the origin. That is a simple pole with residue $2$, since $\lim_{w\to 0}w\cot w=1$.
Welcome to MSE, by the way.

Answer 2

With $t=\dfrac{\pi}{2}-z$ we have $$\dfrac{\cos z}{1-\sin z}=\dfrac{\sin t}{1-\cos t}=\dfrac{t-\dfrac{t^3}{3!}+\cdots}{1-1+\dfrac{t^2}{2!}+\cdots}=\dfrac{1-\dfrac{t^2}{3!}+\cdots}{t\left(\dfrac{1}{2!}-\dfrac{t^2}{4!}+\cdots\right)}$$ so there is a simple pole with residue $2$ there.