Answer
$-2+\sqrt 3$
Work Step by Step
Here, we have $\tan (\dfrac{-\pi}{12})=\tan (\dfrac{2 \pi}{12}-\dfrac{3 \pi}{12})$
This gives: $\tan (\dfrac{2 \pi}{12}-\dfrac{3 \pi}{12})=\tan (\dfrac{\pi}{6}-\dfrac{\pi}{4})$
or, $\dfrac{\tan \dfrac{\pi}{6} -\tan \dfrac{\pi}{4}}{1+ \tan \dfrac{\pi}{6} \tan \dfrac{\pi}{4}}=\dfrac{\dfrac{\sqrt 3}{3}-1}{1+\dfrac{\sqrt 3}{3}\cdot 1}$
or, $ \dfrac{\sqrt 3-3}{\sqrt 3+3} \cdot \dfrac{\sqrt 3-3}{\sqrt 3 -3}=\dfrac{3-6\sqrt 3+9}{3-9}$
Thus, $\tan (\dfrac{-\pi}{12})=-2+\sqrt 3$
