Prove: sin3A/sinA - cos3A/cosA= 2
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L.H.S.
=(Sin 3A)/Sin A-(cos 3A)/cos A
=(3sin A-4sin^3 A)/sin A-(4cos^3 A-3cos A)/cos A
=3-4sin^2 A-4cos^2 A+3
=6-4(sin^2A+cos^A)
=6-4×1
=6-4
=2. Proved
=(Sin 3A)/Sin A-(cos 3A)/cos A
=(3sin A-4sin^3 A)/sin A-(4cos^3 A-3cos A)/cos A
=3-4sin^2 A-4cos^2 A+3
=6-4(sin^2A+cos^A)
=6-4×1
=6-4
=2. Proved
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