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Question
Prove that `sqrt3 "cosec" 20^circ - sin 20^circ` = 4
Solution
LHS = `sqrt3 "cosec" 20^circ - sin 20^circ`
`= sqrt3 * 1/(sin 20^circ) - 1/(cos 20^circ)`
`= (sqrt3 cos 20^circ - sin 20^circ)/(sin 20^circ cos 20^circ)`
`= 2 [(sqrt3/2 cos 20^circ - 1/2 sin 20^circ)/(sin 20^circ cos 20^circ)]`
`= 2 (sin 60^circ cos 20^circ - cos 60^circ sin 20^circ)/(sin 20^circ cos 20^circ)`
`[because sin 60^circ = sqrt3/2 and cos 60^circ = 1/2]`
`= 2 (sin (60^circ - 20^circ))/(sin 20^circ cos 20^circ)`
`= (2 sin 40^circ)/(sin 20^circ cos 40^circ)`
`= (4 sin 40^circ)/(2 sin 20^circ cos 20^circ)`
`= (4 sin 40^circ)/(sin 40^circ)`
[∵ 2 sin A cos A = sin 2A]
= 4 = RHS
Hence proved.
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