prove that sin3@/sin@-cos3@/cos@=2
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L.H.S
=sin3A/sinA - cos3A/cosA
= 3sinA - 4 sin³A / sinA - 4cos³A-3cosA/cosA
=sinA(3-4sin²A)/sinA - cosA(4cos²A-3)/cosA
=3-4sin²A-(4cos²A-3)
= 3-4sin²A-4cos²A+3
=6-4(sin²A+cos²A)
=6-4
=2
=R.H.S
=sin3A/sinA - cos3A/cosA
= 3sinA - 4 sin³A / sinA - 4cos³A-3cosA/cosA
=sinA(3-4sin²A)/sinA - cosA(4cos²A-3)/cosA
=3-4sin²A-(4cos²A-3)
= 3-4sin²A-4cos²A+3
=6-4(sin²A+cos²A)
=6-4
=2
=R.H.S
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