I tried for a while to classify the singularities of $\frac{1}{z}-\frac{1}{\sin z}$ at the origin, but I am stucked
A way to do this it's to consider a hint of a colleague :
$\sin(z)=z\bigl(1+z^2g(z)\bigr)$ where $g$ is analytic with $g(0)\ne0$.
So $$\frac1z-\frac1{\sin z}=\frac{\sin z-z}{z\sin z}=\frac{z^3g(z)}{z^2\bigl(1+z^2g(z)\bigr)}.$$
Is anyone could explain to me in details what he did?
At $z$ close to $0$, $\sin z = z - \frac{z^3}{6} + O(z^4)$.
Thus: $\frac{1}{z} - \frac{1}{\sin z} = \frac{\sin z - z}{z \sin z} = \frac{-z^3/6+O(z^4)}{z^2-z^4/6 + O(z^5)} = \frac{-z/6+O(z^2)}{1-z^2/2+O(z^3)}$.
When $z \to 0$, the above expression converges to $0$. Thus, $0$ is a removable singularity.