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Question
Prove that:
tan 20° tan 40° tan 80° = `sqrt3`.
Solution
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`
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