If circumcentre of an equilateral triangle inscribed in x2a2+y2b2=1 with vertices having eccentric angles a, α,β,γrespectively is (x1,y1) then Σcos⁡αcos⁡β+Σsin⁡αsin⁡β=

centroid x1,y1≡aΣcos⁡α3,bΣsin⁡α3∴ Σcos⁡α=3x1a----1 and Σsin⁡α=3y1b----2Squaring and adding, we getΣcos⁡αcos⁡β+Σsin⁡αsin⁡β=9x122a2+9y122b2−32

If circumcentre of an equilateral triangle inscribed in x2a2+y2b2=1 with vertices having eccentric angles a, α,β,γrespectively is (x1,y1) then Σcos⁡αcos⁡β+Σsin⁡αsin⁡β=

centroid x1,y1≡aΣcos⁡α3,bΣsin⁡α3∴ Σcos⁡α=3x1a----1 and Σsin⁡α=3y1b----2Squaring and adding, we getΣcos⁡αcos⁡β+Σsin⁡αsin⁡β=9x122a2+9y122b2−32

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If circumcentre of an equilateral triangle inscribed in $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with vertices having eccentric angles a, $α, β, γ$
respectively is (x₁,y₁) then $Σcos αcos β + Σsin αsin β =$

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$9 x_{1}^{2} - 9 y_{1}^{2} + a^{2} b^{2}$

$\frac{9 x_{1}^{2}}{2 a^{2}} + \frac{9 y_{1}^{2}}{2 b^{2}} - \frac{3}{2}$

$\frac{9 x_{1}^{2}}{a^{2}} + \frac{9 y_{1}^{2}}{b^{2}} + 3$

$\frac{9 x_{1}^{2}}{a^{2}} + \frac{9 y_{1}^{2}}{b^{2}} + \frac{3}{2}$

answer is C.

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$\begin{array}{l} centroid (x_{1}, y_{1}) \equiv (\frac{a Σcos α}{3}, \frac{b Σsin α}{3}) \\ ∴ Σcos α = \frac{3 x_{1}}{a} - - - - (1) \\ and Σsin α = \frac{3 y_{1}}{b} - - - - (2) \end{array}$

Squaring and adding, we get
$Σcos α \cos β + Σsin α \sin β = \frac{9 x_{1}^{2}}{2 a^{2}} + \frac{9 y_{1}^{2}}{2 b^{2}} - \frac{3}{2}$

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If circumcentre of an equilateral triangle inscribed in x2a2+y2b2=1 with vertices having eccentric angles a, α,β,γrespectively is (x1,y1) then Σcos⁡αcos⁡β+Σsin⁡αsin⁡β=

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If circumcentre of an equilateral triangle inscribed in $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with vertices having eccentric angles a, $α, β, γ$
respectively is (x₁,y₁) then $Σcos αcos β + Σsin αsin β =$