prove that sin3A/sinA - cos3A/cosA =2
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Heya dear
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LHS
sin3A / sinA - cos3A / cosA
= ( sin3A ) ( cosA ) - ( cos3A ) ( sinA ) / sinAcosA
= ( 3sinA - 4sin³A ) cosA - ( 4cos³A - 3cosA ) ( sinA ) / sinAcosA
= 3sinAcosA - 4sin³cosA - 4cos³AsinA + 3sinAcosA / sinAcosA
= 6sinAcosA - 4sin³cosA - 4cos³sinA / sinAcosA
= sinAcosA ( 6 - 4sin²A - 4cos²A ) / sinAcosA
= 6 - 4sin²A - 4cos²A
= 6 - 4 ( sin²A + cos²A ) [ °.° sin²a + cos²b = 1
= 6 - 4 × 1
= 6 - 4
= 2
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