Question

Prove that : (sin θ-2sin^3)/(2cos^3θ-cosθ) = tan θ

Answer 1

we have to prove that, (sinθ - 2sin³θ)/(2cos³θ - cosθ) = tanθ

proof : LHS = (sinθ - 2sin³θ)/(2cos³θ - cosθ)

= {sinθ(1 - 2sin²θ)}/{cosθ(2cos²θ - 1)}

we know, cos2x = cos²x - sin²x

= cos²x - (1 - cos²x)

= 2cos²x - 1

or, cos2x = cos²x - sin²x

= 1 - sin²x - sin²x = 1 - 2sin²x

so, 1 - 2sin²θ = cos2θ .....(1)

2cos²θ - 1 = cos2θ .....(2)

= {sinθ(1 - 2sin²θ)}/{cosθ(2cos²θ - 1)}

from equation (1) and (2) we get,

= {sinθ(cos2θ)}/{cosθ(cos2θ)}

= tanθ = RHS

hence proved//

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