Answer
$-\dfrac{\sqrt {15}}{3}$
Work Step by Step
We have: $\sin a=\dfrac{-\sqrt {15}}{4}$ and $\cos a=\dfrac{-1}{4}$
We plug these values into the following identity:
$\tan (\dfrac{a}{2})=\dfrac{1-\cos a}{\sin a} \\ =\dfrac{1-(\dfrac{-1}{4})}{ (\dfrac{-\sqrt {15}}{4})} \\=-\dfrac{5}{\sqrt {15}}\\=-\dfrac{\sqrt {15}}{3}$
