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प्रश्न
If \[x, y \in \mathbb{R}\], then the determinant
विकल्प
\[\left[ - \sqrt{2}, \sqrt{2} \right]\]
\[\left[ - 1, 1 \right]\]
\[\left[ - \sqrt{2}, 1 \right]\]
\[\left[ - 1, - \sqrt{2} \right]\]
उत्तर
\[∆ = \begin{vmatrix}\cos x & - \sin x & 1 \\ \sin x & \cos x & 1 \\ \cos\left( x + y \right) & - \sin\left( x + y \right) & 0\end{vmatrix}\]
\[ = \begin{vmatrix}\cos x & - \sin x & 1 \\ \sin x & \cos x & 1 \\ 0 & 0 & \sin y - \cos y\end{vmatrix} \left[ \text{Applying }R_3 \to R_3 - \cos y R_1 + \sin y R_2 \right]\]
\[ = \left( \sin y - \cos y \right)\left( \cos^2 x + \sin^2 x \right)\]
\[ = \sin y - \cos y\]
\[ = \sqrt{2}\left( \frac{1}{\sqrt{2}}\sin y - \frac{1}{\sqrt{2}}\cos y \right)\]
\[ = \sqrt{2}\left( \cos\frac{\pi}{4}\sin y - \sin\frac{\pi}{4}\cos y \right)\]
\[ = \sqrt{2}\sin\left( y - \frac{\pi}{4} \right)\]
\[\text{ Therefore,} - \sqrt{2} \leq ∆ \leq \sqrt{2} .\]
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