## Precalculus (6th Edition)

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