[nooonuii]

Let $g(z)=\dfrac{f(z)}{\sin{z}}$. $g$ is analytic at every point except at the points $z=n\pi,n\in\mathbb{Z}$. Wlog assume $z=0$. Since $f(0)=0$ but $f'(0)\neq 0$ (why?) we may define $f(z)=zh(z)$ where $h(z)$ is analytic and $h(0)\neq 0$. Similarly, $\sin{z}=zk(z)$ for some analytic function $k(z)$ where $k(0)\neq 0$. Can we say that $g(z)$ is analytic at $0$?