Answer
$$\tan\frac{\pi}{12}=2-\sqrt3$$
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 $\frac{\pi}{3}$.
$$\tan\frac{\pi}{12}$$
Now, we try write $\pi$ as the difference of $4\pi$ and $3\pi$.
$$\tan\frac{\pi}{12}=\tan \Big(\frac{4\pi-3\pi}{12}\Big)=\tan\Big(\frac{4\pi}{12}-\frac{3\pi}{12}\Big)=\tan\Big(\frac{\pi}{3}-\frac{\pi}{4}\Big)$$
We then use the sine difference identity:
$$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$$
Therefore,
$$\tan\frac{\pi}{12}=\frac{\tan \frac{\pi}{3}-\tan\frac{\pi}{4}}{1+\tan\frac{\pi}{3}\tan\frac{\pi}{4}}$$
$$\tan\frac{\pi}{12}=\frac{\sqrt3-1}{1+\sqrt3\times1}$$
$$\tan\frac{\pi}{12}=\frac{\sqrt3-1}{1+\sqrt3}$$
$$\tan\frac{\pi}{12}=\frac{(\sqrt3-1)^2}{(\sqrt3+1)(\sqrt3-1)}$$ (multiply both numerator and denominator by $\sqrt3-1$)
$$\tan\frac{\pi}{12}=\frac{3+1-2\sqrt3}{3-1}$$
$$\tan\frac{\pi}{12}=\frac{4-2\sqrt3}{2}$$
$$\tan\frac{\pi}{12}=2-\sqrt3$$
