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Rajan@2021
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Prove that: sin^6 θ + cos^6 θ = 1-3sin²θcos²θ

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This question is from trigonometry topic – trigonometric identities in which we have to prove that

sin^6 θ + cos^6 θ = 1-3sin²θcos²θ

RS Aggarwal, Class 10, chapter 13A, question no 17(i)

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1 Answer

  1. L.H.S. =sin6θ+cos6θ
    =(sin²θ+(cos²θ

    Put sin²θ=a and cos²θ=b

     L.H.S. =+
    =(a+b2ab(a+b)
    =(sin²θcos²θ3sin²θcos²θ(sin²θ+cos²θ)
    =(13sin²θcos²θ[sin²θ+cos²θ=1]
    =13sin²θcos²θ
    = R.H.S.

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