simplify and evaluate $\frac{\tan80^\circ-\tan20^\circ}{1+\tan80^\circ\tan20^\circ}$ How do you simplify and evaluate $\dfrac{\tan80^\circ-\tan20^\circ}{1+\tan80^\circ\tan20^\circ}$?
What is the problem asking?
 A: Try the tangent difference formula $$\tan(x-y) = \frac{\tan(x) - \tan(y)}{1+\tan(x)\tan(y)}$$
Also important to remember are these identities:
$$\tan(x\pm y) = \frac{\tan(x) \pm \tan(y)}{1\mp\tan(x)\tan(y)}$$
$$\sin(x\pm y) = \sin(x)\cos(y)\pm \cos(x)\sin(y)$$
$$\cos(x \pm y) = \cos(x)\cos(y) \mp \sin(x)\sin(y)$$
A: 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 $$\frac{\tan 80^\circ-\tan 20^\circ}{1+\tan 80^\circ\tan 20^\circ}$$  $$=\frac{\frac{\sin 80^\circ}{\cos 80^\circ}-\frac{\sin 20^\circ}{\cos 20^\circ}}{1+\frac{\sin 80^\circ}{\cos 80^\circ}\frac{\sin 20^\circ}{\cos 20^\circ}}$$  $$=\frac{\sin 80^\circ\cos 20^\circ-\cos 20^\circ\sin 20^\circ}{\cos 80^\circ\cos 20^\circ+\sin 80^\circ\sin 20^\circ}$$  $$=\frac{\sin(80^\circ-20^\circ)}{\cos(80^\circ-20^\circ)}$$$$=\frac{\sin 60^\circ}{\cos 60^\circ}$$ $$=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}$$ $$ =\color{blue}{\sqrt{3}}$$
A: Hint:
Try to reduce this problem to one that makes use of the more familiar $\sin$ and $\cos$ sum of angle identities.
$$\begin{array}{lll}
\displaystyle\frac{\tan 80^\circ - \tan 20^\circ}{1+\tan80^\circ\tan20^\circ} &=&\displaystyle\frac{\tan 80^\circ + \tan (-20^\circ)}{1-\tan80^\circ\tan(-20^\circ)}\\
&=&\displaystyle\frac{\tan 80^\circ + \tan (-20^\circ)}{1-\tan80^\circ\tan(-20^\circ)}\cdot\frac{\cos 80^\circ\cos(-20^\circ)}{\cos 80^\circ\cos(-20^\circ)}\\
&=&\dots
\end{array}$$
Hint 2: $\tan 80^\circ\cos 80^\circ = ?$ 
You can take it from here I think.
