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Question

If 2tan2α tan2β tan2γ+tan2αtan2β+tan2βtan2γ+tan2γtan2α=1 (where, α,β,γ(2n+1)π2,nπ,nϵI) then

A
sin2α+sin2β+sin2γ=1
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B
cos2α+cos2β+cos2γ=1
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C
cosec2αcosec2β+cosec2βcosec2γ+cosec2γcosec2α=cosec2αcosec2βcosec2γ
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D
cos2α+cos2β+cos2γ=2
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Solution

The correct options are
A sin2α+sin2β+sin2γ=1
C cosec2αcosec2β+cosec2βcosec2γ+cosec2γcosec2α=cosec2αcosec2βcosec2γ
D cos2α+cos2β+cos2γ=2
Dividing by tan2α tan2β tan2γ, we get
2+cot2γ+cot2α+cot2β=cot2αcot2βcot2γ
1+cosec2α=π(cosec2α1)
1+cosec2α=1+cosec2αcosec2α.cosec2β+π(cosec2α)
cosec2α.cosec2β=π(cosec2α)
sin2α+sin2β+sin2γ=1.

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