सारणिक left| begin{array}{ccc} 1 & cos(alpha | vedclass

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(D) माना $D = \left| \begin{array}{ccc} 1 & \cos(\alpha - \beta) & \cos \alpha \ \cos(\alpha - \beta) & 1 & \cos \beta \ \cos \alpha & \cos \beta &

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(D) माना $D = \left| \begin{array}{ccc} 1 & \cos(\alpha - \beta) & \cos \alpha \ \cos(\alpha - \beta) & 1 & \cos \beta \ \cos \alpha & \cos \beta & 1 \end{array} ight|$.प्रथम पंक्ति के अनुदिश विस्तार करने पर:$D = 1(1 - \cos^2 \beta) - \cos(\alpha - \beta)(\cos(\alpha - \beta) - \cos \alpha \cos \beta) + \cos \alpha(\cos \beta \cos(\alpha - \beta) - \cos \alpha)$.सर्वसमिका $\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$ का उपयोग करने पर:$D = \sin^2 \beta - \cos(\alpha - \beta)(\sin \alpha \sin \beta) + \cos \alpha(\cos \beta(\cos \alpha \cos \beta + \sin \alpha \sin \beta) - \cos \alpha)$.$D = \sin^2 \beta - \cos(\alpha - \beta)\sin \alpha \sin \beta + \cos \alpha(\cos \alpha \cos^2 \beta + \sin \alpha \sin \beta \cos \beta - \cos \alpha)$.$D = \sin^2 \beta - \cos(\alpha - \beta)\sin \alpha \sin \beta + \cos^2 \alpha \cos^2 \beta + \sin \alpha \sin \beta \cos \alpha \cos \beta - \cos^2 \alpha$.$D = \sin^2 \beta - \cos(\alpha - \beta)\sin \alpha \sin \beta + \cos^2 \alpha(\cos^2 \beta - 1) + \sin \alpha \sin \beta \cos \alpha \cos \beta$.$D = \sin^2 \beta - \sin \alpha \sin \beta(\cos(\alpha - \beta) - \cos \alpha \cos \beta) - \cos^2 \alpha \sin^2 \beta$.$D = \sin^2 \beta - \sin \alpha \sin \beta(\sin \alpha \sin \beta) - \cos^2 \alpha \sin^2 \beta$.$D = \sin^2 \beta - \sin^2 \alpha \sin^2 \beta - \cos^2 \alpha \sin^2 \beta$.$D = \sin^2 \beta - \sin^2 \beta(\sin^2 \alpha + \cos^2 \alpha) = \sin^2 \beta - \sin^2 \beta(1) = 0$.

सारणिक $\left| \begin{array}{ccc} 1 & \cos(\alpha - \beta) & \cos \alpha \\ \cos(\alpha - \beta) & 1 & \cos \beta \\ \cos \alpha & \cos \beta & 1 \end{array} \right|$ का मान है

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