Solution
= Solution
The zeros of the denominator satisfy $e^z=\ell\pi$ with nonzero integer $\ell$, since an exponential never vanishes. Equivalently the <poles> are
$$
z_{m,k}=\log(m\pi)+ik\pi,\qquad m\geq1,\quad k\in\mathbb Z.
$$
At such a point $e^{z_{m,k}}=(-1)^km\pi$, and the derivative of the denominator is $e^z\cos(e^z)=(-1)^{m+k}m\pi\ne0$. All these <poles> are simple, with
$$
\boxed{\operatorname{Res}_{z_{m,k}}\frac1{\sin(e^z)}=\frac{(-1)^{m+k}}{m\pi}.}
$$