So the question is formulated as follows.
Given the analytic function $z \mapsto f(z) = \dfrac{1}{\sin z} - \dfrac{\cos z}{z}$,
Is $z = 0$ a pole, an essential singularity, a removable singularity, or none of these?
Is it possible, by prescribing a suitable value for $f(0)$, for $z \mapsto f(z)$ to be made holomorphic at $z = 0$ ?
Give the radius of convergence $r$ for the power series $z\mapsto f(z)=\sum_{n = 0}^\infty a_n z^n$, and find the first four terms of this power series.
For the mathematician this would probably take a couple of minutes, if you can give me the final answer I will try to find the steps myself if the steps are too long to write down. I hope someone can answer it, thanks !
I can answer it, but instead I'm going to get you started. Let's look at $1/\sin(z)$. Writing out the Taylor series for $\sin$, this is $$ \frac{1}{\sin z} = \frac{1}{z - z^3/3! + ...} \\ = \frac{1}{z}\frac{1}{(1 - z^2/3! + )} \\ = \frac{1}{z} (1 + z^2/3! + \text{higher order terms}) $$ which looks a lot like $1/z$ near $z = 0$. That has a simple pole at zero.
Does the other term cancel that pole? If so, you might need to look in detail at the higher order terms. I suspect that looking up to the quadratic term will probably suffice.
Addition following comments:
The second term is $$ \frac{\cos z}{z} = \frac{1}{z} - \frac{z}{2!} + \frac{z^3}{4!} - \frac{z^5}{6!} - \ldots $$
Subtracting this from the first term (which I'll re-write out) $$ \frac{1}{\sin z} = \frac{1}{z} (1 + \frac{z^2}{3!} + \text{higher order terms})\\ = \frac{1}{z} + \frac{z}{3!} + \text{higher order terms} $$ gives us $$ f(z) = \frac{1}{z} + \frac{z}{3!} + \text{higher order terms} - (\frac{1}{z} - \frac{z}{2!} + \frac{z^3}{4!} - \frac{z^5}{6!} - \ldots) \\ = \frac{1}{z} + \frac{z}{3!} - \frac{1}{z} + \frac{z}{2!} - \text{higher order terms}\\ = \frac{5z}{6} + \text{higher order terms} $$ So $z = 0$ is a removable singularity. You can "fix" it by defining $f(0) = 0$. The resulting function is holomorphic.
I only wrote out one term of the power series, and I might have gotten the algebra wrong, too. :) I leave it to you to write out the remaining ones, find a formula for the coefficient $a_n$ in terms of $n$, and perform the ratio test. I anticipate that the radius of convergence will turn out to be about $\pi/2$, or maybe $\pi$, but don't trust me on that!