The value of left| {begin{array}{*{20}{c}}1&{ | vedclass

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(B) Let $\Delta = \left| {\begin{array}{*{20}{c}}1&{\cos (\beta - \alpha )}&{\cos (\gamma - \alpha )}\{\cos (\alpha - \beta )}&1&{\cos (\gamma - \bet

Vedclass · Accepted Answer

$\left| {\begin{array}{*{20}{c}}{\sin \alpha }&{\cos \alpha }&0\{\sin \beta }&{\cos \beta }&0\{\sin \gamma }&{\cos \gamma }&0\end{array}} ight|^2$

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(B) Let $\Delta = \left| {\begin{array}{*{20}{c}}1&{\cos (\beta - \alpha )}&{\cos (\gamma - \alpha )}\{\cos (\alpha - \beta )}&1&{\cos (\gamma - \beta )}\{\cos (\alpha - \gamma )}&{\cos (\beta - \gamma )}&1\end{array}} ight|$.Using the identity $\cos(x-y) = \cos x \cos y + \sin x \sin y$,we can write each element as a dot product of two vectors:$\Delta = \left| {\begin{array}{*{20}{c}}{\cos^2 \alpha + \sin^2 \alpha}&{\cos \alpha \cos \beta + \sin \alpha \sin \beta}&{\cos \alpha \cos \gamma + \sin \alpha \sin \gamma}\{\cos \beta \cos \alpha + \sin \beta \sin \alpha}&{\cos^2 \beta + \sin^2 \beta}&{\cos \beta \cos \gamma + \sin \beta \sin \gamma}\{\cos \gamma \cos \alpha + \sin \gamma \sin \alpha}&{\cos \gamma \cos \beta + \sin \gamma \sin \beta}&{\cos^2 \gamma + \sin^2 \gamma}\end{array}} ight|$.This determinant can be expressed as the product of two matrices:$\Delta = \left| {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }&0\{\cos \beta }&{\sin \beta }&0\{\cos \gamma }&{\sin \gamma }&0\end{array}} ight| 	imes \left| {\begin{array}{*{20}{c}}{\cos \alpha }&{\cos \beta }&{\cos \gamma}\{\sin \alpha }&{\sin \beta }&{\sin \gamma}\{0}&{0}&{0}\end{array}} ight|$.Since the second determinant is the transpose of the first,we have:$\Delta = \left| {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }&0\{\cos \beta }&{\sin \beta }&0\{\cos \gamma }&{\sin \gamma }&0\end{array}} ight|^2 = \left| {\begin{array}{*{20}{c}}{\sin \alpha }&{\cos \alpha }&0\{\sin \beta }&{\cos \beta }&0\{\sin \gamma }&{\cos \gamma }&0\end{array}} ight|^2$ (by swapping columns).Thus,the correct option is $B$.

The value of $\left| {\begin{array}{*{20}{c}}1&{\cos (\beta - \alpha )}&{\cos (\gamma - \alpha )}\\{\cos (\alpha - \beta )}&1&{\cos (\gamma - \beta )}\\{\cos (\alpha - \gamma )}&{\cos (\beta - \gamma )}&1\end{array}} \right|$ is

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