Answer
$-2+\sqrt{3}$
Work Step by Step
Use the addition formula for tangent, $\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$.
$\tan\left(\frac{11\pi}{12}\right)$
$=\tan\left(\frac{\pi}{4}+\frac{2\pi}{3}\right)$
$=\frac{\tan\frac{\pi}{4}+\tan\frac{2\pi}{3}}{1-\tan\frac{\pi}{4}\tan\frac{2\pi}{3}}$
$=\frac{1+(-\sqrt{3})}{1-1*(-\sqrt{3})}$
$=\frac{1-\sqrt{3}}{1+\sqrt{3}}$
$=\frac{1-\sqrt{3}}{1+\sqrt{3}}*\frac{1-\sqrt{3}}{1-\sqrt{3}}$
$=\frac{1-2\sqrt{3}+3}{1-3}$
$=\frac{4-2\sqrt{3}}{-2}$
$=-2+\sqrt{3}$
