mathematics 10th trigonometry please solve it fast
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This is Difficult . to now sorry I am in 7:grade
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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.
hope it helps you
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