Answer
$$\tan\frac{11\pi}{12}=\sqrt3-2$$
Work Step by Step
$$X=\tan\frac{11\pi}{12}$$
$11\pi$ in the numerator can be written as the difference of $12\pi$ and $\pi$
$$X=\tan\Big(\frac{12\pi-\pi}{12}\Big)=\tan\Big(\pi-\frac{\pi}{12}\Big)$$
The identity of the difference of tangent states that
$$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$$
Therefore, $$X=\frac{\tan\pi-\tan\frac{\pi}{12}}{1+\tan\pi\tan\frac{\pi}{12}}$$
$$X=\frac{0-\tan\frac{\pi}{12}}{1+0\times\tan\frac{\pi}{12}}$$
$$X=\frac{-\tan\frac{\pi}{12}}{1}$$
$$X=-\tan\frac{\pi}{12}$$
The job remains as to calculating $\tan\frac{\pi}{12}$.
We have $$\tan\frac{\pi}{12}=\tan\Big(\frac{4\pi-3\pi}{12}\Big)=\tan\Big(\frac{\pi}{3}-\frac{\pi}{4}\Big)$$
We apply again identity of the difference of tangents.
$$\tan\frac{\pi}{12}=\frac{\tan\frac{\pi}{3}-\tan\frac{\pi}{4}}{1+\tan\frac{\pi}{3}\tan\frac{\pi}{4}}$$
$$\tan\frac{\pi}{12}=\frac{\sqrt3-1}{1+\sqrt3\times1}$$
$$\tan\frac{\pi}{12}=\frac{\sqrt3-1}{\sqrt3+1}$$
$$\tan\frac{\pi}{12}=\frac{(\sqrt3-1)^2}{(\sqrt3+1)(\sqrt3-1)}$$
$$\tan\frac{\pi}{12}=\frac{3+1-2\sqrt3}{3-1}$$
$$\tan\frac{\pi}{12}=\frac{4-2\sqrt3}{2}$$
$$\tan\frac{\pi}{12}=2-\sqrt3$$
Therefore,
$$X=-\tan\frac{\pi}{12}=\sqrt3-2$$
In conclusion, $$\tan\frac{11\pi}{12}=\sqrt3-2$$
