## Trigonometry (11th Edition) Clone

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