For all values of `theta in(0,pi/2),` the determinant of the matrix `[(-2,tantheta+sec^2theta,3),(-sintheta,costheta,sintheta),(-3,-4,3)]` always lies in the interval :

Options

A. `[(7)/(2), (21)/(4)]`
B. [3, 5]
C. (4, 6)
D. `((5)/(2), (19)/(4))`

Explanation

To find the determinant of the matrix

\[ \begin{bmatrix} -2 & \tan \theta + \sec^2 \theta & 3 \\ -\sin \theta & \cos \theta & \sin \theta \\ -3 & -4 & 3 \end{bmatrix} \]

for all values of \(\theta\) in the interval \((0, \frac{\pi}{2})\), we can use the formula for the determinant of a \(3 \times 3\) matrix. The determinant can be calculated as follows:

### Step 1: Write the determinant formula
The determinant of a \(3 \times 3\) matrix

\[ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \]

is given by:

\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

### Step 2: Substitute the values
For our matrix, we have:

- \(a = -2\)
- \(b = \tan \theta + \sec^2 \theta\)
- \(c = 3\)
- \(d = -\sin \theta\)
- \(e = \cos \theta\)
- \(f = \sin \theta\)
- \(g = -3\)
- \(h = -4\)
- \(i = 3\)

Substituting these values into the determinant formula:

\[ D = -2(\cos \theta \cdot 3 - \sin \theta \cdot (-4)) - (\tan \theta + \sec^2 \theta)(-\sin \theta \cdot 3 - \sin \theta \cdot (-3)) + 3(-\sin \theta \cdot -4 - \cos \theta \cdot (-3)) \]

### Step 3: Simplify each term
Calculating each term step-by-step:

1. **First term:**
\[ -2(3 \cos \theta + 4 \sin \theta) = -6 \cos \theta - 8 \sin \theta \]

2. **Second term:**
\[ -(\tan \theta + \sec^2 \theta)(0) = 0 \]

3. **Third term:**
\[ 3(4 \sin \theta + 3 \cos \theta) = 12 \sin \theta + 9 \cos \theta \]

Combining these results, we have:

\[ D = -6 \cos \theta - 8 \sin \theta + 12 \sin \theta + 9 \cos \theta \]

### Step 4: Combine like terms
Combining like terms gives:

\[ D = (9 \cos \theta - 6 \cos \theta) + (-8 \sin \theta + 12 \sin \theta) = 3 \cos \theta + 4 \sin \theta \]

### Step 5: Determine the range of \(D\)
To find the range of \(D = 3 \cos \theta + 4 \sin \theta\) for \(\theta \in (0, \frac{\pi}{2})\), we can express it in the form \(R \sin(\theta + \phi)\):

1. Calculate \(R\):
\[ R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

2. The maximum value of \(D\) is \(R = 5\) and the minimum value occurs when \(\theta\) is such that \(3 \cos \theta + 4 \sin \theta\) is minimized.

3. The minimum value occurs when \(\theta\) is such that \(\tan \phi = \frac{4}{3}\), which gives the minimum value of \(D\) as \(0\) when \(\theta = 0\).

### Step 6: Final interval
Thus, the determinant \(D\) lies in the interval \((0, 5)\) for \(\theta \in (0, \frac{\pi}{2})\).

### Conclusion
The determinant of the matrix always lies in the interval \((3, 5)\).