simplify and evaluate $\frac{\tan80^\circ-\tan20^\circ}{1+\tan80^\circ\tan20^\circ}$ How do you simplify and evaluate $\dfrac{\tan80^\circ-\tan20^\circ}{1+\tan80^\circ\tan20^\circ}$?
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Method 1: Notice formula, $$\color{red}{\frac{\tan A-\tan B}{1+\tan A\tan B}=\tan (A-B)}$$ Hence, we have $$\frac{\tan 80^\circ-\tan 20^\circ}{1+\tan 80^\circ\tan 20^\circ}$$$$=\tan (80^\circ-20^\circ)$$ $$ =\tan (60^\circ)$$ $$ =\color{blue}{\sqrt{3}}$$
Method 2: Notice, $$\color{red}{\sin A\cos B-\cos A\sin B=\sin (A-B)}$$ & $$\color{red}{\cos A\cos B+\sin A\sin B=\cos (A-B)}$$  Now, we have $$\frac{\tan 80^\circ-\tan 20^\circ}{1+\tan 80^\circ\tan 20^\circ}$$  $$=\frac{\frac{\sin 80^\circ}{\cos 80^\circ}-\frac{\sin 20^\circ}{\cos 20^\circ}}{1+\frac{\sin 80^\circ}{\cos 80^\circ}\frac{\sin 20^\circ}{\cos 20^\circ}}$$  $$=\frac{\sin 80^\circ\cos 20^\circ-\cos 20^\circ\sin 20^\circ}{\cos 80^\circ\cos 20^\circ+\sin 80^\circ\sin 20^\circ}$$  $$=\frac{\sin(80^\circ-20^\circ)}{\cos(80^\circ-20^\circ)}$$$$=\frac{\sin 60^\circ}{\cos 60^\circ}$$ $$=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}$$ $$ =\color{blue}{\sqrt{3}}$$
