We know that,
sinA-sinB=2cos\( ( \dfrac{A+B}{2})sin( \dfrac{A-B}{2})\)
And cos2A-sin2A=cos2A
L.H.S. = \( \dfrac{sinx-sin3x}{sin^2x-cos^2x}\)
= \( \dfrac{2cos(\dfrac{x+3x}{2})sinx(\dfrac{x-3x}{2})}{-cos2x}\)
= -2× (-sinx)
Therefore , we have,
L.H.S=2sinx
= R.H.S.
Hence proved.
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