Question

Prove that sin theta - 2 sin ^3 theta / 2 cos ^3 theta - cos theta = tan theta.. guys please stop answering just for the sake of points...

Answer 1

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Step-by-step explanation:

(Sinθ  - 2Sin³θ) / ( 2Cos³θ - Cosθ)  = Tanθ

LHS = (Sinθ  - 2Sin³θ) / ( 2Cos³θ - Cosθ)

= Sinθ( 1 - 2Sin²θ) /Cosθ( 2Cos²θ - 1)

1 = Cos²θ + Sin²θ

= Sinθ(  Cos²θ + Sin²θ - 2Sin²θ) /Cosθ( 2Cos²θ - ( Cos²θ + Sin²θ))

= Sinθ(  Cos²θ  - Sin²θ) /Cosθ( 2Cos²θ -  Cos²θ - Sin²θ)

= Sinθ(  Cos²θ  - Sin²θ) /Cosθ( Cos²θ  - Sin²θ)

Cancelling  Cos²θ  - Sin²θ from numerator & denominator

= Sinθ /Cosθ

= Tanθ

= RHS

QED

proved

(Sinθ  - 2Sin³θ) / ( 2Cos³θ - Cosθ)  = Tanθ

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