Question

Fools easiest problem of the day,

Prove that \( \tan(20^\circ) \cdot \tan(40^\circ) \cdot \tan(80^\circ)= \tan(60^\circ). \)

Answer 1

Lets take \(x=20\), it's easier with variables.
\[\tan x \cdot \tan (60-x)\cdot \tan(60+x)\] \[\implies -\tan x \cdot \frac{\tan x -\tan 60}{1 + \tan x \tan 60}\cdot \frac{\tan x +\tan 60}{1 - \tan x \tan 60}\]\[\implies -\tan x \cdot \frac{\tan^2x - 3}{1 - 3 \tan^2 x} =- \frac{\tan^3x - 3\tan x}{1 - 3\tan^2x} = \tan3x \text{ or,}\quad \tan 60\]