Prove that: sin theta - 2sin ^3 theta / 2cos ^ 3 theta - cos theta = tan theta.
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LHS
(sinθ - 2 sin³θ)/(2cos³θ - cosθ)
=[sinθ(1 - 2 sin²θ)]/[cosθ(2cos²θ - 1)]
=[sinθ(1 - sin²θ- sin²θ)]/[cosθ(cos²θ+cos²θ - 1)]
∵1 - sin²θ = cos²θ
and cos²θ-1 = - sin²θ
∴=[sinθ(1 - sin²θ- sin²θ)]/[cosθ(cos²θ+cos²θ - 1)]
=[sinθ(cos²θ- sin²θ)]/[cosθ(cos²θ-sin²θ)]
= sinθ/cosθ = tanθ = RHS
(sinθ - 2 sin³θ)/(2cos³θ - cosθ)
=[sinθ(1 - 2 sin²θ)]/[cosθ(2cos²θ - 1)]
=[sinθ(1 - sin²θ- sin²θ)]/[cosθ(cos²θ+cos²θ - 1)]
∵1 - sin²θ = cos²θ
and cos²θ-1 = - sin²θ
∴=[sinθ(1 - sin²θ- sin²θ)]/[cosθ(cos²θ+cos²θ - 1)]
=[sinθ(cos²θ- sin²θ)]/[cosθ(cos²θ-sin²θ)]
= sinθ/cosθ = tanθ = RHS
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