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Question

Expression sin6θcos6θsin2θcos2θ is equal to

A
sin4θcos4θ
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B
1sin2θcos2θ
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C
1+sin2θcos2θ
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D
13sin2θcos2θ
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Solution

The correct option is B 1sin2θcos2θ
sin6θcos6θsin2θcos2θ=(sin2θ)3(cos2θ)3sin2θcos2θ
=(sin2θcos2θ)(sin4θ+cos4θ+sin2θcos2θ)(sin2θcos2θ)
=[(sin2θ+cos2θ)22sin2θ×cos2θ+sin2θcos2θ]
=1sin2θcos2θ.

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