Question

PROOF THAT sin3A /sinA-cos3A/cosA=2

Answer 1

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

RHS

thanks

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Answer 2

Hey there!

Refer the picture for your answer.

Answer in attachment .