#### Answer

$$\frac{\tan80^\circ+\tan55^\circ}{1-\tan80^\circ\tan55^\circ}=-1$$

#### Work Step by Step

$$X=\frac{\tan80^\circ+\tan55^\circ}{1-\tan80^\circ\tan55^\circ}$$
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
$$X=\tan(80^\circ+55^\circ)$$
$$X=\tan135^\circ$$
As we know from Section 5.3:
$$\tan\theta=\cot(90^\circ-\theta)$$
So, $$X=\tan135^\circ=\cot(90^\circ-135^\circ)=\cot(-45^\circ)$$
Also, from Negative-Angle Identities:
$$\cot(-\theta)=-\cot\theta$$
Therefore, $$X=-\cot45^\circ$$
$$X=-1$$
Overall, $$\frac{\tan80^\circ+\tan55^\circ}{1-\tan80^\circ\tan55^\circ}=-1$$