Let tanα,tanβ,tanγ ;α,β,γ,γ≠(2n-1)π2,n∈N be the slopes of three line segments OA,OB and OC respectively, where O is origin. If Circum center of ΔABC coincides with origin and its ortho center lies on y-axis, the value of cos3α+cos3β+cos3γcosαcosβcosγ2 is equal to
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answer is 144.
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Detailed Solution
A=(cosα,sinα),B=(cosβ,sinβ),C=(cosγ,sinγ) Circum center =(0,0)Centroid G divides the HS in the ratio 2:1 internally Ortho center H=(cosα+cosβ+cosγ,sinα+sinβ+sinγ) lies on y -axis ⇒cosα+cosβ+cosγ=0⇒cos3α+cos3β+cos3γ=3cosαcosβcosγ∑cos3α=4∑cos3α-3∑cosα=12cosαcosβcosγ+0 Given expression =12cosαcosβcosγcosαcosβcosγ2=144
Let tanα,tanβ,tanγ ;α,β,γ,γ≠(2n-1)π2,n∈N be the slopes of three line segments OA,OB and OC respectively, where O is origin. If Circum center of ΔABC coincides with origin and its ortho center lies on y-axis, the value of cos3α+cos3β+cos3γcosαcosβcosγ2 is equal to