Answer
$2-\sqrt 3$
Work Step by Step
Double -angle Theorem can be defined as: $\tan \dfrac{\theta}{2}=\pm \sqrt {\dfrac{1- \cos \theta}{1+ \cos \theta}}$
$\tan (-165 ^{\circ})=\tan \dfrac{(-330^{\circ})}{2}$
Since $\tan$ is negative in the second quadrant.
Thus, $- \sqrt {\dfrac{1- \cos (-330^{\circ})}{1+ \cos (-330^{\circ})}}= \sqrt {\dfrac{1- \cos (360-330^{\circ})}{1+ \cos (360^{\circ}+330 ^{\circ})}}=\sqrt {\dfrac{1- \cos 30 ^{\circ}}{1+ \cos 30^{\circ}}}$
and $\sqrt {\dfrac{1- (\sqrt 3/2)}{1+(\sqrt 3/2)}}=\sqrt {\dfrac{2-\sqrt 3}{2+\sqrt 3}}=\sqrt {\dfrac{2-\sqrt 3}{2+\sqrt 3}} \times \sqrt {\dfrac{2-\sqrt 3}{2-\sqrt 3}}$
Hence, $\tan (-165 ^{\circ})=2-\sqrt 3$
