#### Answer

$$\frac{\tan80^\circ-\tan(-55^\circ)}{1+\tan80^\circ\tan(-55^\circ)}=-1$$

#### Work Step by Step

$$X=\frac{\tan80^\circ-\tan(-55^\circ)}{1+\tan80^\circ\tan(-55^\circ)}$$
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 $A=80^\circ$ and $B=-55^\circ$.
Therefore, $X$ can also be rewritten as
$$X=\tan[80^\circ-(-55^\circ)]$$
$$X=\tan(80^\circ+55^\circ)$$
$$X=\tan135^\circ$$
We can rewrite $135^\circ$ as the difference of $180^\circ$ and $45^\circ$
$$X=\tan(180^\circ-45^\circ)$$
and apply the identity of tangent difference:
$$X=\frac{\tan180^\circ-\tan45^\circ}{1+\tan180^\circ\tan45^\circ}$$
$$X=\frac{0-1}{1+0\times1}$$
$$X=\frac{-1}{1}=-1$$
Overall, $$\frac{\tan80^\circ-\tan(-55^\circ)}{1+\tan80^\circ\tan(-55^\circ)}=-1$$