Answer
$1$
Work Step by Step
Recall:
$\tan{(\theta+180^{\circ})} = \tan{\theta}$
This means that
$\tan{200^{\circ}} = \tan{(20^{\circ}+180^{\circ})} = \tan{20^{\circ}}$
Thus,
$\tan{200^{\circ}} \cdot \cot{20^{\circ}} = \tan{20^{\circ}}\cdot \cot{20^{\circ}}$
Since $\cot{\theta} = \dfrac{1}{\tan{\theta}}$, then
$\cot{20^{\circ}} = \dfrac{1}{\tan{20^{\circ}}}$
Therefore,
$\tan{20^{\circ}}\cdot \cot{20^{\circ}} = \tan{20^{\circ}}\cdot \dfrac{1}{\tan{20^{\circ}}} = \boxed{1}$
