Answer
$$\tan285^\circ=-2-\sqrt3$$
Work Step by Step
$$X=\tan285^\circ$$
$285^\circ$ can be written as the difference of $360^\circ$ and $75^\circ$ (do not write it as the sum of $270^\circ$ and $15^\circ$ since $\tan270^\circ$ is undefined).
$$X=\tan(360^\circ-75^\circ)$$
The identity of the difference of tangent states that
$$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$$
Therefore, $$X=\frac{\tan 360^\circ-\tan 75^\circ}{1+\tan360^\circ\tan75^\circ}$$
$\tan360^\circ$ is actually $\tan0^\circ=0$.
$$X=\frac{0-\tan75^\circ}{1+0\times\tan75^\circ}$$
$$X=\frac{-\tan75^\circ}{1}$$
$$X=-\tan75^\circ$$
The job remains as to calculating $\tan75^\circ$.
We can rewrite $75^\circ$ as the sum of $45^\circ$ and $30^\circ$ and apply the identities of tangent sum:
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
$$\tan75^\circ=\tan(45^\circ+30^\circ)=\frac{\tan45^\circ+\tan30^\circ}{1-\tan45^\circ\tan30^\circ}$$
$$\tan75^\circ=\frac{1+\frac{1}{\sqrt3}}{1-1\times\frac{1}{\sqrt3}}=\frac{\frac{\sqrt3+1}{\sqrt3}}{1-\frac{1}{\sqrt3}}=\frac{\frac{\sqrt3+1}{\sqrt3}}{\frac{\sqrt3-1}{\sqrt3}}=\frac{\sqrt3+1}{\sqrt3-1}$$
$$\tan75^\circ=\frac{(\sqrt3+1)^2}{(\sqrt3-1)(\sqrt3+1)}=\frac{3+1+2\sqrt3}{3-1}=\frac{4+2\sqrt3}{2}=2+\sqrt3$$
Therefore,
$$X=-\tan75^\circ=-2-\sqrt3$$
In conclusion, $$\tan285^\circ=-2-\sqrt3$$
