Prove That : Sinx - Sin3xsin^2x - Cos^2x = 2sin X - Toppr

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SolveGuidesJoin / LoginUse appLogin0You visited us 0 times! Enjoying our articles? Unlock Full Access!Standard XIMathsTrigonometric Functions of Sum and Difference of Two anglesQuestionProve that :$$\dfrac{\sin x -\sin 3x}{\sin^2x -\cos^2x}=2\sin x$$Open in AppSolutionVerified by Toppr

LHS :$$\dfrac{\sin x -\sin 3x}{\sin^2x-\cos^2x}$$$$=\dfrac{2\sin (\dfrac{x-3x}{2}) \cos (\dfrac{x+3x}{2})}{-(\cos^2x -\sin^2x)}$$$$=\dfrac{2\sin (-x) \cos 2x}{-\cos 2x}$$$$=\dfrac{-2\sin x \cos 2x}{-\cos 2x}$$$$=2\sin x$$ = RHS

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