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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L.H.S. =sin6θ+cos6θ
=(sin²θ)³+(cos²θ)³
Put sin²θ=a and cos²θ=b
∴ L.H.S. =a³+b³
=(a+b)³−2ab(a+b)
=(sin²θ−cos²θ)³−3sin²θcos²θ(sin²θ+cos²θ)
=(1)³−3sin²θcos²θ[∵sin²θ+cos²θ=1]
=1−3sin²θcos²θ
= R.H.S.