Answer
$$\tan\frac{5\pi}{12}=2+\sqrt3$$
Work Step by Step
*STRATEGY: In this type of exercise, we would try to rewrite the angle in terms of already familiar angles, including $\frac{\pi}{6}$, $\frac{\pi}{4}$ and $\frac{\pi}{3}$.
$$\tan\frac{5\pi}{12}$$
Now, we try writing $5\pi$ as the sum of $2\pi$ and $3\pi$.
$$\tan\frac{5\pi}{12}=\tan \Big(\frac{2\pi+3\pi}{12}\Big)=\tan\Big(\frac{2\pi}{12}+\frac{3\pi}{12}\Big)=\tan\Big(\frac{\pi}{6}+\frac{\pi}{4}\Big)$$
We then use the tangent sum identity:
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
Therefore,
$$\tan\frac{5\pi}{12}=\frac{\tan \frac{\pi}{6}+\tan\frac{\pi}{4}}{1-\tan\frac{\pi}{6}\tan\frac{\pi}{4}}$$
$$\tan\frac{5\pi}{12}=\frac{\frac{1}{\sqrt3}+1}{1-\frac{1}{\sqrt3}\times1}$$
$$\tan\frac{5\pi}{12}=\frac{\frac{1+\sqrt3}{\sqrt3}}{1-\frac{1}{\sqrt3}}$$
$$\tan\frac{5\pi}{12}=\frac{\frac{\sqrt3+1}{\sqrt3}}{\frac{\sqrt3-1}{\sqrt3}}$$
$$\tan\frac{5\pi}{12}=\frac{\sqrt3+1}{\sqrt3-1}$$
Multiply both numerator and denominator with $\sqrt3+1$
$$\tan\frac{5\pi}{12}=\frac{(\sqrt3+1)^2}{(\sqrt3-1)(\sqrt3+1)}$$
$$\tan\frac{5\pi}{12}=\frac{3+1+2\sqrt3}{3-1}$$
$$\tan\frac{5\pi}{12}=\frac{4+2\sqrt3}{2}$$
$$\tan\frac{5\pi}{12}=2+\sqrt3$$