Prove that $ \tan 20°\cdot \tan 40° \cdot \tan 60° \cdot \tan 80° = 3$

Question

We have to prove that $$ \tan 20°\cdot \tan 40° \cdot \tan 60° \cdot \tan 80° = 3$$

I tried to use the formula

$$\displaystyle\frac {\tan a+\tan b}{1-\tan a\cdot \tan b}$$

But in that I got stuck .

Answer 1

Hint:

$$\tan\alpha\tan(60^{\circ}-\alpha)\tan(60^{\circ}+\alpha)=\tan3\alpha$$ If $\alpha=20^{\circ}$, then $$\tan20^{\circ}\tan40^{\circ}\tan80^{\circ}=\tan60^{\circ}=\sqrt3$$