If alpha+beta+gamma=2 pi, then the system of | vedclass

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Question

$\alpha+\beta+\gamma=2 \pi$

$\left|\begin{array}{ccc}1 & \cos \gamma & \cos \beta \ \cos \gamma & 1 & \cos \alpha \ \cos \beta &a

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Vedclass · Answer

$\alpha+\beta+\gamma=2 \pi$

$\left|\begin{array}{ccc}1 & \cos \gamma & \cos \beta \ \cos \gamma & 1 & \cos \alpha \ \cos \beta & \cos \alpha & 1\end{array}ight|$

$=1+2 \cos \alpha \cdot \cos \beta \cdot \cos \gamma-\cos ^{2} \alpha-\cos ^{2} \beta-\cos ^{2} \gamma$

$=\sin ^{2} \gamma-\cos ^{2} \alpha-\cos ^{2} \beta+(\cos (\alpha+\beta)+\cos (\alpha-\beta)) \cos \gamma$

$=\sin ^{2} \gamma-\cos ^{2} \alpha-\cos ^{2} \beta+\cos ^{2} \gamma+\cos (\alpha-\beta) \cos \gamma$

$=\sin ^{2} \alpha-\cos ^{2} \beta+\cos (\alpha-\beta) \cos (\alpha+\beta)$

$=\sin ^{2} \alpha-\cos ^{2} \beta+\cos ^{2} \alpha-\sin ^{2} \beta=0$

If $\alpha+\beta+\gamma=2 \pi$, then the system of equations

$x+(\cos \gamma) y+(\cos \beta) z=0$

$(\cos \gamma) x+y+(\cos \alpha) z=0$

$(\cos \beta) x+(\cos \alpha) y+z=0$

has :

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