Answer
$${\text{Undefined}}$$
Work Step by Step
$$\eqalign{
& \cot \left( { - 13\pi } \right) \cr
& {\text{Use }}\cot \left( { - \theta } \right) = - \cot \theta \cr
& \cot \left( { - 13\pi } \right) = - \cot \left( {13\pi } \right) \cr
& \cr
& {\text{Write }}13\pi {\text{ as 6}}\left( {{\text{2}}\pi } \right) + \pi \cr
& \cot \left( { - 13\pi } \right) = - \cot \left( {{\text{6}}\left( {{\text{2}}\pi } \right) + \pi } \right) \cr
& \cot \left( { - 13\pi } \right) = - \cot \left( \pi \right) \cr
& \cr
& {\text{Where }}\cot \theta = \frac{{\cos \theta }}{{\sin \theta }} \cr
& \cot \left( { - 13\pi } \right) = - \frac{{\cos \pi }}{{\sin \pi }} \cr
& \cr
& {\text{Simplify}} \cr
& \cot \left( { - 13\pi } \right) = \frac{1}{0}{\text{ Undefined}} \cr} $$
