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**Prove It** Since this is a simple pole, the residue can be calculated by $\displaystyle \displaystyle \begin{align*} \textrm{Res}\, (f, c) = \lim_{ z \to c } (z - c) f(z) \end{align*}$. So in this case where $\displaystyle \displaystyle \begin{align*} f(z) = \frac{1}{\sin{z}} \end{align*}$ we have

$\displaystyle \displaystyle \begin{align*} \textrm{Res}\,(f,\pi) &= \lim_{z \to \pi} \frac{z - \pi}{\sin{z}} \\ &= \lim_{z \to \pi}\frac{1}{\cos{z}} \textrm{ by L'Hospital's Rule} \\ &= \frac{1}{\cos{\pi}} \\ &= \frac{1}{-1} \\ &= -1 \end{align*}$