Answer
$$\tan\Big(-\frac{7\pi}{12}\Big)=2+\sqrt3$$
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 $\frac{\pi}{3}$.
$$\tan\Big(-\frac{7\pi}{12}\Big)$$
From Negative-Angle Identities:
$$\tan(-\theta)=-\tan\theta$$
So that means, $$\tan\Big(-\frac{7\pi}{12}\Big)=-\tan\frac{7\pi}{12}$$
Now, we write $7\pi$ as the sum of $3\pi$ and $4\pi$.
$$\tan\frac{7\pi}{12}=\tan \Big(\frac{4\pi+3\pi}{12}\Big)=\tan\Big(\frac{4\pi}{12}+\frac{3\pi}{12}\Big)=\tan\Big(\frac{\pi}{3}+\frac{\pi}{4}\Big)$$
Therefore, $$\tan\Big(-\frac{7\pi}{12}\Big)=-\tan\Big(\frac{\pi}{3}+\frac{\pi}{4}\Big)$$
We then use the tangent sum identity:
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
Therefore,
$$\tan\Big(-\frac{7\pi}{12}\Big)=-\frac{\tan\frac{\pi}{3}+\tan\frac{\pi}{4}}{1-\tan\frac{\pi}{3}\tan\frac{\pi}{4}}$$
$$\tan\Big(-\frac{7\pi}{12}\Big)=-\frac{\sqrt3+1}{1-\sqrt3\times1}$$
$$\tan\Big(-\frac{7\pi}{12}\Big)=-\frac{1+\sqrt3}{1-\sqrt3}$$
We multiply both numerator and denominator by $1+\sqrt3$
$$\tan\Big(-\frac{7\pi}{12}\Big)=-\frac{(1+\sqrt3)^2}{(1-\sqrt3)(1+\sqrt3)}$$
$$\tan\Big(-\frac{7\pi}{12}\Big)=-\frac{1+3+2\sqrt3}{1-3}$$
$$\tan\Big(-\frac{7\pi}{12}\Big)=-\frac{4+2\sqrt3}{-2}$$
$$\tan\Big(-\frac{7\pi}{12}\Big)=\frac{4+2\sqrt3}{2}$$
$$\tan\Big(-\frac{7\pi}{12}\Big)=2+\sqrt3$$
