Answer
$-2-\sqrt 3$
Work Step by Step
Here, we have $\tan (\dfrac{7\pi}{12})=\tan (\dfrac{3 \pi}{12}+\dfrac{4 \pi}{12})$
This gives: $\tan (\dfrac{3 \pi}{12}+\dfrac{4 \pi}{12})=\tan (\dfrac{\pi}{4}+\dfrac{\pi}{3})$
or, $\dfrac{\tan \dfrac{\pi}{4} +\tan \dfrac{\pi}{3}}{1 - \tan \dfrac{\pi}{4} \tan \dfrac{\pi}{3}}=\dfrac {1+\sqrt 3}{1-(1)(\sqrt 3)}$
or, $ \dfrac{1+\sqrt 3}{1-\sqrt 3} \cdot \dfrac{1+\sqrt 3}{1+\sqrt 3}=\dfrac{1+2\sqrt 3+3}{1-3}$
Thus, $\tan (\dfrac{7\pi}{12})=-2-\sqrt 3$
