Question
$\left| {\,\begin{array}{*{20}{c}}1&{\cos (\beta - \alpha )}&{\cos (\gamma - \alpha )}\\{\cos (\alpha - \beta )}&1&{\cos (\gamma - \beta )}\\{\cos (\alpha - \gamma )}&{\cos (\beta - \gamma )}&1\end{array}} \right|$ का मान होगा
$ = \,\left| {\,\begin{array}{*{20}{c}}{{{\cos }^2}\alpha + {{\sin }^2}\alpha }&{\cos \beta \cos \alpha + \sin \beta \sin \alpha }&{\cos \alpha \cos \gamma + \sin \alpha \sin \gamma }\\{\cos \alpha \cos \beta + \sin \alpha \sin \beta }&{{{\cos }^2}\beta + {{\sin }^2}\beta }&{\cos \beta \cos \gamma + \sin \beta \sin \gamma }\\{\cos \alpha \cos \gamma + \sin \alpha \sin \gamma }&{\cos \beta \cos \gamma + \sin \beta \sin \gamma }&{{{\cos }^2}\beta + {{\sin }^2}\beta }\end{array}\,} \right|$
$ = \,\left| {\,\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }&0\\{\cos \beta }&{\sin \beta }&0\\{\cos \gamma }&{\sin \gamma }&0\end{array}\,} \right|\,.\,\left| {\,\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }&0\\{\cos \beta }&{\sin \beta }&0\\{\cos \gamma }&{\sin \gamma }&0\end{array}\,} \right| = {\left| {\,\begin{array}{*{20}{c}}{\sin \alpha }&{\cos \alpha }&0\\{\sin \beta }&{\cos \beta }&0\\{\sin \gamma }&{\cos \gamma }&0\end{array}\,} \right|^2}$.
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