If the minimum and the maximum values of the function f: [(π)/(4),...

${C}_{2} \to {C}_{2} - {C}_{1}$

f\left(\theta\right) = \left|\begin{cases}- \sin^{2}\theta & - 1 & 1 \\ - \cos^{2}\theta & - 1 & 1 \\ 12 & - 2 & - 2\end{cases}\right| = 4\left(\cos^{2}\theta - \sin^{2}\theta\right) = 4\left(\cos 2\theta\right)

Since $\cos 2\theta = \cos^{2}\theta - \sin^{2}\theta$

Again, $\dfrac{\pi}{4} \le \theta \le \dfrac{\pi}{2} ⇒ 2\left(\dfrac{\pi}{4}\right) \le 2\theta \le 2\left(\dfrac{\pi}{2}\right)$

$⇒ \dfrac{\pi}{2} \le \theta \le \pi$$∴ \theta \in \left[\dfrac{\pi}{4}, \dfrac{\pi}{2}\right] ⇒ 2\theta \in \left[\dfrac{\pi}{2}, \pi\right]$ i.e., Second Quadrant.

$⇒ - 1 \le \cos 2\theta \le 0$

f\left(\theta\right){}_\max = M = 0

f{\left(\theta\right)}_\min = m = - 4