1. We are asked to find the residue of the function $f(z) = \frac{1}{\sin z}$ at its singularity points. 2. The function $\sin z$ has zeros at $z = n\pi$ for all integers $n$. 3. These points are the singularities of $f(z)$ because $f(z)$ has poles where the denominator vanishes. 4. To find the residue of $f(z)$ at $z = n\pi$, we recall that near a simple pole $z_0$, the residue of $f(z)$ is given by $$\text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0) f(z).$$ 5. In our case, $f(z) = \frac{1}{\sin z}$ has simple poles at $z = n\pi$ because $\sin z$ has simple zeros there. 6. Using the limit definition: $$\text{Res}(f, n\pi) = \lim_{z \to n\pi} (z - n\pi) \frac{1}{\sin z} = \lim_{z \to n\pi} \frac{z - n\pi}{\sin z}.$$ 7. Using the fact that $\sin z$ near $z = n\pi$ behaves like: $$\sin z \approx (-1)^n (z - n\pi),$$ we substitute in the limit to get $$\text{Res}(f, n\pi) = \lim_{z \to n\pi} \frac{z - n\pi}{(-1)^n (z - n\pi)} = \frac{1}{(-1)^n} = (-1)^n.$$ 8. Therefore, the residue of $1/\sin z$ at $z = n\pi$ is $(-1)^n$. Final answer: $$\boxed{\text{Res}(f, n\pi) = (-1)^n}.$$