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Question

The value of sin6θ+cos6θ+3sin2θcos2θ is

A
0
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B
1
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C
2
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D
3
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Solution

The correct option is B 1
We have
sin6θ+cos6θ+3sin2θcos2θ=sin6θ+cos6θ+3sin2θcos2θ(sin2θ+cos2θ)
=(sin2θ+cos2θ)3=1

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