Answer
$-\sqrt {3} $
Work Step by Step
Period of $cot$ function is $\pi$
It means
$cot\left( \alpha \right) =cot\left( \alpha +\pi k\right) =cot\left( \alpha -\pi k\right) $
$k$ is integer number
$\cot \left( -390\right) =\cot \left( -360-30\right) =\cot \left( -2\pi -\dfrac {\pi }{6}\right) =\cot \left( -\dfrac {\pi }{6}\right) =\dfrac {\cos \left( 0-\dfrac {\pi }{6}\right) }{\sin \left( 0-\dfrac {\pi }{6}\right) }=\dfrac {\cos 0\cos \dfrac {\pi }{6}+\sin _{0}\sin \dfrac {\pi }{6}}{\sin 0\cos \dfrac {\pi }{6}-\cos 0\sin \dfrac {\pi }{6}}=\dfrac {\cos \dfrac {\pi }{6}+0}{0-\sin \dfrac {\pi }{6}}= =-\dfrac {\cos \dfrac {\pi }{6}}{\sin \dfrac {\pi }{6}}=-\dfrac {\dfrac {\sqrt {3}}{2}}{\dfrac {1}{2}}=-\sqrt {3} $
