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7 .Prove that sin® - 2sin’0e WY no.2¢05°6 - ८०58 |
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Answer» (sinθ- 2 sin^3θ)/(2cos^3θ-cosθ) =sinθ(1-2sin^2θ)/cosθ(2cos^2 θ-1) = sin θ [1- 2(1- cos^2 θ)/cos θ(2cos ^2 θ -1) = sin θ [1-2+ 2cos ^2 θ]/ cos θ (2 cos^2θ -1) = sin θ(2cos^2 θ -1)/ cos θ(2 cos ^2θ-1) = sin θ/ cos θ = tanθ Like my answer if you find it useful! |
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