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

1 Answer

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

Related Questions

Evalute the following:

(i) \( \dfrac{sin 20°}{ cos 70°}\)

(ii) \(\dfrac{ cos 19°}{ sin 71°}\)

(iii) \( \dfrac{sin 21°}{ cos 69°}\)

(iv) \( \dfrac{tan 10°}{ cot 80°}\)

(v) \( \dfrac{sec 11°}{ cosec 79°}\)

Class 10 Maths Introduction to Trigonometry View Answer