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Step-by-Step Solution
Step 1: Write down the given function
We have
$$
f(\theta) =
\begin{vmatrix}
-\sin^2\theta & -1 - \sin^2\theta & 1 \\
-\cos^2\theta & -1 - \cos^2\theta & 1 \\
12 & 10 & -2
\end{vmatrix},
\quad
\text{for } \theta \in \left[\frac{\pi}{4}, \frac{\pi}{2}\right].
$$
Step 2: Apply column operations to simplify the determinant
Operation 1:
$C_1 \to C_1 - C_2$ and $C_3 \to C_3 + C_2$.
This transforms the determinant into
$$
\begin{vmatrix}
1 & -1 - \sin^2\theta & -\sin^2\theta \\
1 & -1 - \cos^2\theta & -\cos^2\theta \\
2 & 10 & 8
\end{vmatrix}.
$$
Operation 2:
$C_2 \to C_2 - C_3$.
This leads to
$$
\begin{vmatrix}
1 & -1 & -\sin^2\theta \\
1 & -1 & -\cos^2\theta \\
2 & 2 & 8
\end{vmatrix}.
$$
Step 3: Expand the resulting determinant
Let us expand along the first row for clarity:
$$
f(\theta) =
1 \cdot
\begin{vmatrix}
-1 & -\cos^2\theta \\
2 & 8
\end{vmatrix}
\;-\;
(-1)\cdot
\begin{vmatrix}
1 & -\cos^2\theta \\
2 & 8
\end{vmatrix}
\;+\;
\bigl(-\sin^2\theta\bigr)\cdot
\begin{vmatrix}
1 & -1 \\
2 & 2
\end{vmatrix}.
$$
Simplifying step by step:
First minor:
$
\begin{vmatrix}
-1 & -\cos^2\theta \\
2 & 8
\end{vmatrix}
= -1 \cdot 8 - (-\cos^2\theta) \cdot 2 = -8 + 2\cos^2\theta.
$
Second minor:
$
\begin{vmatrix}
1 & -\cos^2\theta \\
2 & 8
\end{vmatrix}
= 1 \cdot 8 - (-\cos^2\theta)\cdot 2 = 8 + 2\cos^2\theta.
$
Third minor:
$
\begin{vmatrix}
1 & -1 \\
2 & 2
\end{vmatrix}
= 1 \cdot 2 - (-1)\cdot 2 = 2 + 2 = 4.
$
Putting it all together:
$$
f(\theta)
= 1\bigl(-8 + 2\cos^2\theta\bigr)
\;-\;(-1)\bigl(8 + 2\cos^2\theta\bigr)
\;+\;\bigl(-\sin^2\theta\bigr)(4).
$$
After careful simplification (and noticing from the given solution steps that intermediate manipulations further reduce to $4\cos(2\theta)$), one obtains:
$$
f(\theta) = 4\cos(2\theta).
$$
Step 4: Determine the range of $\cos(2\theta)$ over the given interval
Since $\theta \in \left[\frac{\pi}{4}, \frac{\pi}{2}\right],$ it follows that
$$
2\theta \in \left[\frac{\pi}{2}, \pi\right].
$$
Over this interval, $\cos(2\theta)$ takes values in the interval $[-1,0].$
Hence
$$
f(\theta) = 4\cos(2\theta) \in [4 \cdot (-1),\, 4 \cdot 0] = [-4,\, 0].
$$
Therefore, the minimum value of $f(\theta)$ is $-4$ and the maximum value is $0.$
Step 5: Conclude the ordered pair (m, M)
From the range of $f(\theta)$, we identify:
$$
m = -4 \quad\text{and}\quad M = 0.
$$
Thus,
$$
(m, M) = (-4, 0).
$$
