Question

sin x - sin 3x/sin^2x-cos^2X

Answer 1

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please refer to the attachment above.....

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Answer 2

LHS=(sinx−sin3x)/(sin^2x − cos^2x)

=(sinx−(3sinx−4sin^3x))/(sin^2x−(1−sin^2x))

=(sinx−3sinx+4sin^3x)/(2sin^2x-1)

=(4 sin^3x − 2 sin x)/(2 sin^2x − 1)

=(2 sin x cancel((2 sin^2x − 1)))/cancel((2 sin^2x − 1))

= 2 sin x=RHS