$\begin{vmatrix}\mathbf{1} & \mathbf{\cos \alpha} & \mathbf{1} \\-\sin \alpha & 1 & -\cos \alpha \\-1 & \sin \alpha & 1\end{vmatrix}$
$= [1(1+sin\alpha cos \alpha ) - cos \alpha (-sin \alpha - cos \alpha)+1 (-sin^2 \alpha + 1)]$
$= 1+ sin \alpha cos \alpha) + sin \alpha cos \alpha + cos^2 \alpha - sin^2 \alpha + 1 $
$=2 + 2 sin \alpha cos \alpha + (cos^2 \alpha - sin^2 \alpha) $
$= 2+ sin2 \alpha + cos2 \alpha$
$=2+\sqrt{2} ( cos2\alpha \frac{1}{\sqrt{2}} + sin 2 \alpha\frac{1}{\sqrt{2}} $
$=2+\sqrt{2} (cos (2\alpha - \frac{\pi}{4})) $
તેથી $ -1 \leq cos (2 \alpha - \frac{\pi}{4})\leq 1 $
$\Rightarrow -\sqrt{2}\leq\sqrt{2} cos(2\alpha - \frac{\pi}{4}) \leq \sqrt{2} $
$\Rightarrow2-\sqrt{2} \leq 2 + \sqrt{2} cos (2 \alpha - \frac{\pi}{4}) \leq 2+\sqrt{2} $
$\Rightarrow2-\sqrt{2} \leq D \leq 2+ \sqrt{2} $
$D \in [2 - \sqrt{2}, 2 + \sqrt {2} ] $
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