If origin is the orthocentre of a triangle formed by the options (cosα,sinα,0),(cosβ,sinβ,0),(cosγ,sinγ,0) then ∑cos(2α−β−γ)=−

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answer is D.

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Detailed Solution

OA=OB=OC;G=H=O(0,0,0)Equilateral trianglecosα+cosβ=−cosγ,sinα+sinβ=−sinγ,square and add cos(α−β)=−12cos(β−γ)=cos(γ−α)cos(2α−β−γ)=cos(α−β)−(γ−α)=1

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