## Trigonometry (11th Edition) Clone

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