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Question
Prove the following:
`(sin x - sin 3x)/(sin^2 x - cos^2 x) = 2sin x`
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Solution
We have, L.H.S. = `(sin x - sin 3x)/(sin^2 x - cos^2 x)`
= `(2sin ((x - 3x)/2) cos ((x + 3x)/2))/(-cos^2x - sin^2x)`
= `(2sin (-x) cos (2x))/(-cos2x)`
= `(- 2sin x cos(2x))/(-cos2x)` [∵ cos2x - sin2x = cos 2x]
= `(-2 sinx)/-1` [∵ cos2x - sin2x = cos 2x]
= 2 sin x = R.H.S.
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