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Q.

tan(α+β−γ)tan(α−β+γ)=tanγtanβ, and β≠γ, then the value of sin2α+sin2β+sin2γ is

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a

0

b

1

c

2

d

12

answer is A.

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Detailed Solution

tan(α+β−γ)+tan(α−β+γ)tan(α+β−γ)−tan(α−β+γ)=tanγ+tanβtanγ−tanβ⇒sin2αsin2(β−γ)=sin(γ+β)sin(γ−β)⇒sin2α−2cos(γ−β)=sin(γ+β)⇒sin2α=−2sin(γ+β)cos(γ−β)=−[sin2γ+sin2β]
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