Answer
1.907415
Work Step by Step
Converting $-512^{\circ}20'$ to decimal degrees, we get
$-512^{\circ}20'=-(512^{\circ}+\frac{20}{60}^{\circ})=-(512^{\circ}+0.333333^{\circ})=-512.333333^{\circ}$
$\cot(-512^{\circ}20')=\cot(-512.333333^{\circ})=\frac{1}{\tan (-512.333333^{\circ})}\approx1.907415$
