Question

Prove that (sinA - 2sin^3A)/(2cos^3A-cosA) =tanA​

Answer 1

LHS= sinA-2sin^3 A

2cos^3A-cosA

=sinA(1-sin^2A-sin^2A)

cosA(cos^2A+cos^2A-1)

=sinA(cos^2A-sin^2A)

cos A(cos^2A-sin^2A)

=sinA

cosA

=tanA

=RHS

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Answer 2

$\frac{sina - 2 {sin}^{3} a}{2 {cos}^{3} a - cosa} \\ \\ = \frac{sina(1 - 2 {sin}^{2} a)}{cosa(2 {cos}^{2}a - 1) } \\ \\ = \frac{sina(cos2a)}{cosa(cos2a)} \\ \\ = \frac{sina}{cosa} \\ \\ = tana$