In order to find the Laurent series of $\frac{1}{\sin{x}}$, I used a well-known formula:
$\sum_{n=0}^{\infty}{x^n} = \frac{1}{1-x} \quad (|x| < 1)$
by
$\frac{1}{\sin{x}} = \frac{1}{z}\frac{1}{1+\left(\sum_{n=1}^{\infty}{\frac{(-1)^n}{(2n+1)!}}z^{2n}\right)} = \frac{1}{z}\sum_{m=0}^{\infty}\left(\sum_{n=1}^{\infty}{\frac{(-1)^n}{(2n+1)!}}z^{2n}\right)^m$
I think this answer is true, but I wonder how I can check the condition of the formula: $|x| < 1$ at this time. Would you be able to lend your expertise?
You could first consider the Taylor expansion of $$\frac{\sin(x)}x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n+1)!}$$ So $$\frac x{\sin(x)}=1+\frac{x^2}{6}+\frac{7 x^4}{360}+\frac{31 x^6}{15120}+\frac{127 x^8}{604800}+O\left(x^{10}\right)$$ but the problem is to recognize that the coefficients are $$\frac{(-4)^n }{(2 n)!}B_{2 n}\left(\frac{1}{2}\right)$$ where appears Bernoulli polynomials.