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

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Answer 1 · 2018-02-24T14:50:07

Below is the ans

#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.