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Ex 8.3, 4 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (vii) (sin θ − 2 sin3 θ)/(2 cos3 θ − cos θ)=tan θ Solving L.H.S (sin θ − 2 sin3θ)/(2 cos3 θ − cosθ) = (𝒔𝒊𝒏⁡𝜽 (1 − 2 sin2 θ))/(𝐜𝐨𝐬 𝛉(2cos2 θ − 1)) = sin⁡θ/(cos θ) × ( (1 − 2sin2θ))/((2cos2θ − 1)) = tan θ × ( (1 − 2sin2θ))/((2𝐜𝐨𝐬𝟐𝛉 − 1)) We know that cos2 θ + sin2 θ = 1 cos2 θ = 1 – sin2 θ = tan θ × ((1 − 2sin2θ))/((2(𝟏 − 𝐬𝐢𝐧𝟐𝜽) −1) ) = tan θ × ((1 − 2sin2θ))/((2 − 2sin2θ − 1)) = tan θ × ((𝟏 − 𝟐𝐬𝐢𝐧𝟐𝛉))/((𝟏 − 𝟐𝐬𝐢𝐧𝟐𝛉) ) = tan θ × 1 = tan θ = R.H.S Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo