We are given $$f(z) = \frac{1}{\sin z} - \frac{1}{e^z-1}$$
$\frac{1}{\sin z}$ has poles in $k \pi, \ \ k \in \mathbb{Z}$ and $\frac{1}{e^z-1}$ has singularities in $2 k \pi i, \ \ k \in \mathbb{Z}$.
Here $0$ is the common singularity.
How can I establish the singularities of the difference of those functions $= f(z)?$
$k\,\pi$ for $k\ne0$ and $2\,k\pi\,i$ are order one poles. For $z=0$ we must check what happens, since both singularities could cancel. In fact, $$ \lim_{z\to0}\Bigl(\frac{1}{\sin z} - \frac{1}{e^z-1}\Bigr)=\frac12, $$ so that $z=0$ is a removable singularity.