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(sin θ – 2sin^3θ)/(2cos^3θ-cos θ) = tan θ

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Evaluate:

\(\frac{sin θ – 2sin^3θ}{2cos^3θ-cos θ}\) = tan θ

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\(\frac{sin θ – 2sin^3θ}{2cos^3θ-cos θ}\) = tan θ

L.H.S. = \(\frac{sin θ \; – 2sin^3θ}{2cos^3θ-\;cos θ}\)

Take sin θ as in numerator and cos θ in denominator as outside, it becomes

= [sin θ(1 – 2sin2θ)]/[cos θ(2cos2θ- 1)]

We know that sin2θ = 1-cos2θ

= sin θ[1 – 2(1-cos2θ)]/[cos θ(2cos2θ -1)]

= [sin θ(2cos2θ -1)]/[cos θ(2cos2θ -1)]

= tan θ = R.H.S.

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