Prove that sinx-sin3x/sin^2x-cos^2x=2sinx ?

1 Answer

Below is the ans

Explanation:

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

Hence proved.