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B
1−sin2θcos2θ
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C
1+sin2θcos2θ
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D
1−3sin2θcos2θ
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Solution
The correct option is B1−sin2θcos2θ sin6θ−cos6θsin2θ−cos2θ=(sin2θ)3−(cos2θ)3sin2θ−cos2θ =(sin2θ−cos2θ)(sin4θ+cos4θ+sin2θcos2θ)(sin2θ−cos2θ) =[(sin2θ+cos2θ)2−2sin2θ×cos2θ+sin2θcos2θ] =1−sin2θcos2θ.