Prove the following identities, where the angles involved are acute angles for which the expressions are defined.$$\dfrac{ (sin\; θ – 2sin^3\;θ)}{(2cos^3θ-cos θ)}= tan θ$$

Asked by Abhisek | 1 year ago |  179

##### Solution :-

L.H.S. = $$\dfrac{sin\; θ – 2sin^3\;θ}{2cos^3\;θ-cos\; θ}$$

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

$$\dfrac{sin\; θ(1 – 2sin^2\;θ)}{cos\; θ(2cos^2\;θ- 1})$$

We know that sin2θ = 1-cos2θ

$$\dfrac{ sin θ[1 – 2(1-cos^2θ)]}{[cos θ(2cos^2θ -1)]}$$

$$\dfrac{[sin θ(2cos2θ -1)]}{[cos θ(2cos2θ -1)]}$$

= tan θ = R.H.S.

Answered by Pragya Singh | 1 year ago

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