Let $\sum_{n=-\infty}^\infty a_n z^n$ be the Laurent series of $\frac{1}{\sin (z)}$ for $|z| < \pi$. I'm asked to prove that if $n < -1$ or $n$ is even, then $a_n = 0$. I'm also asked to compute $a_{-1}$ and $a_1$.
What's a good way to approach this problem?
Consider the annulus $A=\{ z: 0< |z| < \pi\}$. Let $f(z)=1/\sin(z)$, since $z=0$ is a simple zero of $1/f$, then $z=0$ is a simple pole for $f$, which gives that $a_{-n}=0$ for all $n>1$. Thus the Laurent series of $f$ in $A$ is
$$
f(z)= a_{-1} \frac{1}{z} + \sum_{n=0}^{\infty}a_n z^n \tag{1}
$$
Of course, since $f$ is an odd function then all the even coefficients must be zero and that is why $a_n=0$ for $n=0,2,4,\cdots$. Finally is easily seen from (1) that
$$
a_{-1} = \lim_{z \to 0} z \ f(z) = \lim_{z \to 0} \frac{z}{\sin(z)} =1 ,
$$
again from (1) and since $a_{-1}=1$, we get
$$
a_1 = \lim_{z\to 0} \left(\frac{f(z)}{z}-\frac{1}{z^2}\right)=\lim_{z\to 0} \frac{z-\sin(z)}{z^2\sin(z)}=\frac{1}{6},
$$
where you can compute the last limit by three consecutive application of L´Hospital's rule.
I do not know if this is very orthodox with respect to the manner this topic is teached to you. So forgive me if I am off-topic.
Consider $$f(z)=\frac{1}{\sin(z)}$$ and the Taylor expansion $$\sin(z)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}z^{2n+1}$$ Perform the long division and get $$f(z)=\frac{1}{z}+\frac{z}{6}+\frac{7 z^3}{360}+\frac{31 z^5}{15120}+O\left(z^6\right)$$
