Answer
$$\tan\Big(-\frac{5\pi}{12}\Big)=-2-\sqrt3$$
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 $\frac{\pi}{3}$.
$$\tan\Big(-\frac{5\pi}{12}\Big)$$
From Negative-Angle Identities:
$$\tan(-\theta)=-\tan\theta$$
So that means, $$\tan\Big(-\frac{5\pi}{12}\Big)=-\tan\frac{5\pi}{12}$$
Now, we write $5\pi$ as the sum of $3\pi$ and $2\pi$.
$$\tan\frac{5\pi}{12}=\tan \Big(\frac{2\pi+3\pi}{12}\Big)=\tan\Big(\frac{2\pi}{12}+\frac{3\pi}{12}\Big)=\tan\Big(\frac{\pi}{6}+\frac{\pi}{4}\Big)$$
Therefore, $$\tan\Big(-\frac{5\pi}{12}\Big)=-\tan\Big(\frac{\pi}{6}+\frac{\pi}{4}\Big)$$
We then use the tangent sum identity:
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
Therefore,
$$\tan\Big(-\frac{5\pi}{12}\Big)=-\frac{\tan\frac{\pi}{6}+\tan\frac{\pi}{4}}{1-\tan\frac{\pi}{6}\tan\frac{\pi}{4}}$$
$$\tan\Big(-\frac{5\pi}{12}\Big)=-\frac{\frac{1}{\sqrt3}+1}{1-\frac{1}{\sqrt3}\times1}$$
$$\tan\Big(-\frac{5\pi}{12}\Big)=-\frac{\frac{1}{\sqrt3}+1}{1-\frac{1}{\sqrt3}}$$
We multiply both numerator and denominator by $1+\frac{1}{\sqrt3}$
$$\tan\Big(-\frac{5\pi}{12}\Big)=-\frac{\Big(\frac{1}{\sqrt3}+1\Big)^2}{\Big(1-\frac{1}{\sqrt3}\Big)\Big(1+\frac{1}{\sqrt3}\Big)}$$
$$\tan\Big(-\frac{5\pi}{12}\Big)=-\frac{\frac{1}{3}+1+\frac{2}{\sqrt3}}{1-\frac{1}{3}}$$
$$\tan\Big(-\frac{5\pi}{12}\Big)=-\frac{\frac{1+3+2\sqrt3}{3}}{\frac{2}{3}}$$
$$\tan\Big(-\frac{5\pi}{12}\Big)=-\frac{\frac{4+2\sqrt3}{3}}{\frac{2}{3}}$$
$$\tan\Big(-\frac{5\pi}{12}\Big)=-\frac{4+2\sqrt3}{2}$$
$$\tan\Big(-\frac{5\pi}{12}\Big)=-2-\sqrt3$$
