Question

Prove that:

tan 20° tan 40° tan 80° = `sqrt3`.

Answer 1

tan 20° tan 40° tan 80°

`= (sin 20^circ)/(cos 20^circ) xx (sin 40^circ)/(cos 40^circ) xx (sin 80^circ)/(cos 80^circ)`

`= (sin 20^circ xx sin 40^circ xx sin 80^circ)/(cos 20^circ cos 40^circ cos 80^circ)`

Consider sin 20° × sin 40° sin 80°

= sin 20° sin (60° – 20°) sin (60° + 20°)

= sin 20° [sin² 60° – sin² 20°]

`= sin 20^circ [3/4 - sin^2 20^circ]`

`= sin 20^circ [(3 - 4 sin^2 20^circ)/4]`

`= (3  sin 20^circ - 4 sin^3 20^circ)/4`

`= (sin  60^circ)/4`

`= (sqrt3/2)/4 = sqrt3/8`

cos 20° × cos 40° cos 80° = `1/8` ....[∵ from (i)] .... (2)

divide (1) by (2) we get, tan 20° tan 40° tan 80° = `(sqrt3/8)/(1/8) = sqrt3`