Answer
$-\frac{\sqrt 3}{3}$
Work Step by Step
$\cot 2280^{\circ}=\frac{1}{\tan 2280^{\circ}}$
As the period of tangent function is $180^{\circ}$,
$\tan 2280^{\circ}=\tan (2280^{\circ}-180^{\circ}\times13)$
$=\tan (-60^{\circ})=-\tan 60^{\circ}=-\sqrt {3}$
$\cot 2280^{\circ}=\frac{1}{-\sqrt 3}=-\frac{\sqrt 3}{3}$
