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Question

The value of sin25212sin22212 is

A
3122
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B
3+122
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C
3142
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D
3+142
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Solution

The correct option is D 3+142
We know that,
sin2Asin2B=sin(A+B)sin(AB)
sin25212sin22212=sin75sin30=sin30sin(45+30)=sin30°(sin45°cos30°+sin30°cos45°)=12×3+122=3+142

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