Answer
$$\frac{\tan100^\circ+\tan80^\circ}{1-\tan100^\circ\tan80^\circ}=0$$
Work Step by Step
$$X=\frac{\tan100^\circ+\tan80^\circ}{1-\tan100^\circ\tan80^\circ}$$
From the identity of the sum of tangent:
$$\frac{\tan A+\tan B}{1-\tan A\tan B}=\tan(A+B)$$
So here $X$ actually follows the above identity with $A=100^\circ$ and $B=80^\circ$.
Therefore, $X$ can also be rewritten as
$$X=\tan(100^\circ+80^\circ)$$
$$X=\tan180^\circ$$
$$X=0$$
which means
$$\frac{\tan100^\circ+\tan80^\circ}{1-\tan100^\circ\tan80^\circ}=0$$
