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Question

If α+β+γ=π, then the value of sin2α+sin2βsin2γ is equal to

A
2sinα
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B
2sinαcosβsinγ
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C
2sinαsinβcosγ
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D
2sinαsinβsinγ
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Solution

The correct option is C 2sinαsinβcosγ
Given, α+β+γ=π
Therefore, sin2α+sin(βγ)sin(β+γ)
=sin2α+sin(πα)sin(βγ)
=sinα[sinα+sin(βγ)]
=sinα[sin(π(β+γ))+sin(βγ)]
=sinα[sin(βγ)+sin(β+γ)]
=sinαsinβcosγ

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