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प्रश्न
\[∆ = \begin{vmatrix}\cos \alpha \cos \beta & \cos \alpha \sin \beta & - \sin \alpha \\ - \sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{vmatrix}\]
उत्तर
Given:
\[∆ = \begin{vmatrix}\cos \alpha \cos \beta & \cos \alpha \sin \beta & - \sin \alpha \\ - \sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{vmatrix}\]
\[\Rightarrow ∆ = \left( - 1 \right)^{1 + 1} \cos\alpha \cos\beta\left( \cos\alpha \cos\beta - 0 \right) + \left( - 1 \right)^{1 + 2} \cos\alpha \sin\beta\left( - \sin\beta \cos\alpha - 0 \right) + \left( - 1 \right)^{1 + 3} \left( - \sin\alpha \right)\left( - \sin^2 \beta \sin\alpha - \sin\alpha \cos^2 \beta \right) \left[ \text{ Expanding along }R_1 \right]\]
\[ = \cos\alpha \cos\beta\left( \cos\alpha \cos\beta - 0 \right) - \cos\alpha \sin\beta\left( - \sin\beta \cos\alpha - 0 \right) - \sin\alpha\left( - \sin^2 \beta \sin\alpha - \sin\alpha \cos^2 \beta \right)\]
\[ = \cos^2 \alpha \cos^2 \beta + \cos^2 \alpha \sin^2 \beta + \sin^2 \alpha \sin^2 \beta + \sin^2 \alpha \cos^2 \beta\]
\[ = \cos^2 \alpha\left( \cos^2 \beta + \sin^2 \beta \right) + \sin^2 \alpha\left( \sin^2 \beta + \cos^2 \beta \right) \]
\[\]
\[ \Rightarrow ∆ = \cos^2 \alpha + \sin^2 \alpha \left[ \because \sin^2 \theta + \cos^2 \theta = 1 \right]\]
\[\]
\[ \Rightarrow ∆ = 1 \left[ \because \sin^2 \theta + \cos^2 \theta = 1 \right]\]
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