SinA-2sin³A/2cos³A-cosA=tanA
Prove that?
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Answer:
L.H.S=SinA-2sin³A/2cos³A-cosA
=sinA(1-2sin²A)/cosA(2cos²A-1)
=sinA(Sin²A+cos²A-sin²A-sin²A)/cosA(cos²A+cos²A-sin²A-cos²A)
(Sins, Sin²A+cos²A=1)
=sinA(cos²A-sin²A)/CosA(cos²A-sin²A)
=sinA/cosA
=TanA
=R.H.S.
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