prove that. sin3A/sinA+cos3A/cosA=4cos2A
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sin3A/sinA+cos3A/cosA
=(sin3AcosA+cos3AsinA)/sinAcosA
=2sin(3A+A)/2sinAcosA [∵, sinαcosβ+cosαsinβ=sin(α+β)]
=2sin4A/sin2A [∵, 2sinαcosα=sin2α]
=(2×2sin2Acos2A)/sin2A
=4cos2A (Proved)
=(sin3AcosA+cos3AsinA)/sinAcosA
=2sin(3A+A)/2sinAcosA [∵, sinαcosβ+cosαsinβ=sin(α+β)]
=2sin4A/sin2A [∵, 2sinαcosα=sin2α]
=(2×2sin2Acos2A)/sin2A
=4cos2A (Proved)
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