left| {,begin{array}{*{20}{c}}1&{cos (bet | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) $\Delta = \,\left| {\,\begin{array}{*{20}{c}}1&{\cos (\beta - \alpha )\,}&{\cos (\gamma - \alpha )}\{\cos (\alpha - \beta )\,}&1&{\cos (\gamma - \beta )}\{\cos (\alpha - \gamma )}&{\cos (\beta - \gamma )\,}&1\end{array}\,} ight|$

$ = \,\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}\,} ight|$

$ = \,\left| {\,\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }&0\{\cos \beta }&{\sin \beta }&0\{\cos \gamma }&{\sin \gamma }&0\end{array}\,} ight|\,.\,\left| {\,\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }&0\{\cos \beta }&{\sin \beta }&0\{\cos \gamma }&{\sin \gamma }&0\end{array}\,} ight| = {\left| {\,\begin{array}{*{20}{c}}{\sin \alpha }&{\cos \alpha }&0\{\sin \beta }&{\cos \beta }&0\{\sin \gamma }&{\cos \gamma }&0\end{array}\,} ight|^2}$.

$\left| {\,\begin{array}{*{20}{c}}1&{\cos (\beta - \alpha )}&{\cos (\gamma - \alpha )}\\{\cos (\alpha - \beta )}&1&{\cos (\gamma - \beta )}\\{\cos (\alpha - \gamma )}&{\cos (\beta - \gamma )}&1\end{array}} \right|$ का मान होगा

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