If the minimum and the maximum values of the function $$f : \left[\frac{\pi}{4}, \frac{\pi}{2}\right] \to R$$, defined by$$f(\theta) = \begin{vmatrix} -\sin^2\theta & -1 - \sin^2\theta & 1 \\ -\cos^2\theta & -1 - \cos^2\theta & 1 \\ 12 & 10 & -2 \end{vmatrix}$$ are $$m$$ and $$M$$ respectively, then the ordered pair $$(m, M)$$ is equal to:

We first observe that the given function is the determinant of a $$3 \times 3$$ matrix that depends on the angle $$\theta$$:

$$ f(\theta)= \begin{vmatrix} -\sin^{2}\theta & -1-\sin^{2}\theta & 1\\[4pt] -\cos^{2}\theta & -1-\cos^{2}\theta & 1\\[4pt] 12 & 10 & -2 \end{vmatrix}. $$

For any square matrix, the determinant formula along the first row is

$$ \begin{vmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{vmatrix} = a_{11} \begin{vmatrix} a_{22}&a_{23}\\ a_{32}&a_{33} \end{vmatrix} - a_{12} \begin{vmatrix} a_{21}&a_{23}\\ a_{31}&a_{33} \end{vmatrix} + a_{13} \begin{vmatrix} a_{21}&a_{22}\\ a_{31}&a_{32} \end{vmatrix}. $$

We now label the first-row entries of our matrix:

$$ a_{11}=-\sin^{2}\theta,\quad a_{12}=-1-\sin^{2}\theta,\quad a_{13}=1. $$

Likewise, the needed sub-determinants are computed one by one.

1. The first cofactor:

$$ \begin{vmatrix} a_{22}&a_{23}\\ a_{32}&a_{33} \end{vmatrix} = \begin{vmatrix} -1-\cos^{2}\theta & 1\\ 10 & -2 \end{vmatrix} = (-1-\cos^{2}\theta)(-2)-1\cdot 10 = 2+2\cos^{2}\theta-10 = -8+2\cos^{2}\theta. $$

Multiplying by $$a_{11}$$ we get

$$ a_{11}\bigl(-8+2\cos^{2}\theta\bigr) = -\sin^{2}\theta\,(2\cos^{2}\theta-8) = -2\sin^{2}\theta\cos^{2}\theta+8\sin^{2}\theta. $$

2. The second cofactor:

$$ \begin{vmatrix} a_{21}&a_{23}\\ a_{31}&a_{33} \end{vmatrix} = \begin{vmatrix} -\cos^{2}\theta & 1\\ 12 & -2 \end{vmatrix} = (-\cos^{2}\theta)(-2)-1\cdot 12 = 2\cos^{2}\theta-12. $$

Multiplying by $$-a_{12}$$ (note the minus sign from the formula) gives

$$ -\bigl(-1-\sin^{2}\theta\bigr)(2\cos^{2}\theta-12) = (1+\sin^{2}\theta)(2\cos^{2}\theta-12). $$

Expanding inside:

$$ (1+\sin^{2}\theta)(2\cos^{2}\theta-12) = 2\cos^{2}\theta+2\sin^{2}\theta\cos^{2}\theta-12-12\sin^{2}\theta. $$

3. The third cofactor:

$$ \begin{vmatrix} a_{21}&a_{22}\\ a_{31}&a_{32} \end{vmatrix} = \begin{vmatrix} -\cos^{2}\theta & -1-\cos^{2}\theta\\ 12 & 10 \end{vmatrix} = (-\cos^{2}\theta)(10)-(-1-\cos^{2}\theta)(12) = -10\cos^{2}\theta+12+12\cos^{2}\theta = 12+2\cos^{2}\theta. $$

Since $$a_{13}=1$$, this term contributes exactly

$$ 12+2\cos^{2}\theta. $$

4. Adding all three contributions:

First contribution:

$$-2\sin^{2}\theta\cos^{2}\theta+8\sin^{2}\theta.$$

Second contribution:

$$2\cos^{2}\theta+2\sin^{2}\theta\cos^{2}\theta-12-12\sin^{2}\theta.$$

Third contribution:

$$12+2\cos^{2}\theta.$$

Now we combine like terms step by step.

• The mixed term $$-2\sin^{2}\theta\cos^{2}\theta$$ from the first line cancels exactly with $$+2\sin^{2}\theta\cos^{2}\theta$$ from the second line.

• Collecting the $$\sin^{2}\theta$$ terms:

$$8\sin^{2}\theta-12\sin^{2}\theta=-4\sin^{2}\theta.$$

• Collecting the $$\cos^{2}\theta$$ terms:

$$2\cos^{2}\theta+2\cos^{2}\theta=4\cos^{2}\theta.$$

• The constants $$-12$$ and $$+12$$ cancel out.

Thus the determinant simplifies neatly to

$$ f(\theta)=4\cos^{2}\theta-4\sin^{2}\theta =4\bigl(\cos^{2}\theta-\sin^{2}\theta\bigr). $$

Recalling the double-angle identity $$\cos2\theta=\cos^{2}\theta-\sin^{2}\theta$$, we rewrite the function as

$$ f(\theta)=4\cos2\theta. $$

Next we examine the interval restriction $$\frac{\pi}{4}\le\theta\le\frac{\pi}{2}$$. Multiplying by 2 gives

$$ \frac{\pi}{2}\le 2\theta\le\pi. $$

On this interval the cosine function decreases steadily from

$$ \cos\!\left(\frac{\pi}{2}\right)=0 \quad\text{down to}\quad \cos(\pi)=-1. $$

Therefore

$$ -1\le\cos2\theta\le 0. $$

Multiplying every part of this inequality by the positive constant 4, we obtain

$$ -4\le f(\theta)\le 0. $$

Thus the minimum value is $$m=-4$$ and the maximum value is $$M=0$$. The required ordered pair is

$$ (m,M)=(-4,0). $$

Among the given alternatives, this matches Option B.

Hence, the correct answer is Option B.