If α, β are the roots of the equation ax^2 +bx+c=0, then the value of the determinant determinant ((1,(cos (β−α)),(cos α)),((cos (α−β)),1,(cos β)),((cos α),(cos β),1)) is


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Question Number 41820 by v tfvhjdxf last updated on 13/Aug/18

$If α, β are the roots of the equation$ ${a x}^{2} + b x + c = 0, then the value of the$ $determinant$ $\| \begin{matrix} 1 & \cos (β - α) & \cos α \\ \cos (α - β) & 1 & \cos β \\ \cos α & \cos β & 1 \end{matrix} \| is$
	Answered by tanmay.chaudhury50@gmail.com last updated on 15/Aug/18
	$1(1−cos^2 β)−cos(β−α){cos(α−β)−cosαcosβ}+cosα{cos(α−β)cosβ−cosα} sin^2 β−cos(β−α)(sinαsinβ)+cosαcosβcos(α−β)−cos^2 α sin^2 β−cos^2 α+cos(α−β){cosαcosβ−sinαsinβ) sin^2 β−cos^2 α+cos(α−β)cos(α+β) (1/2)(2sin^2 β−2cos^2 α+cos2α+cos2β) (1/2)(2sin^2 β−2cos^2 α+2cos^2 α−1+1−2sin^2 β) (1/(2())×0=0 pls check$
	$1 (1 - {c o s}^{2} β) - c o s (β - α) {c o s (α - β) - c o s α c o s β} + c o s α {c o s (α - β) c o s β - c o s α}$ ${s i n}^{2} β - c o s (β - α) (s i n α s i n β) + c o s α c o s β c o s (α - β) - {c o s}^{2} α$ ${s i n}^{2} β - {c o s}^{2} α + c o s (α - β) {c o s α c o s β - s i n α s i n β)$ ${s i n}^{2} β - {c o s}^{2} α + c o s (α - β) c o s (α + β)$ $\frac{1}{2} (2 {s i n}^{2} β - 2 {c o s}^{2} α + c o s 2 α + c o s 2 β)$ $\frac{1}{2} (2 {s i n}^{2} β - 2 {c o s}^{2} α + 2 {c o s}^{2} α - 1 + 1 - 2 {s i n}^{2} β)$ $\frac{1}{2 (} \times 0 = 0$ $p l s c h e c k$
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