Let `Delta=|(1,sin theta, 1),(-sin theta, 1, sin theta),(-1,-sin theta,1)|0,le theta le 2pi`. The

Options

A. `Delta=0`
B. `Delta epsilon (2,oo)`
C. `Delta epsilon (2,4)`
D. `Delta epsilon [2,4]`

Answer

To solve the determinant \( \Delta = \begin{vmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{vmatrix} \), we will calculate the determinant step by step.

### Step 1: Write down the determinant
We start with the determinant:
\[ \Delta = \begin{vmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{vmatrix} \]

### Step 2: Use the determinant formula
The formula for a 3x3 determinant is:
\[ \Delta = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \]

For our matrix:
- \( a = 1, b = \sin \theta, c = 1 \)
- \( d = -\sin \theta, e = 1, f = \sin \theta \)
- \( g = -1, h = -\sin \theta, i = 1 \)

### Step 3: Calculate the determinant
Now we calculate each part:
1. \( ei - fh = 1 \cdot 1 - \sin \theta \cdot (-\sin \theta) = 1 + \sin^2 \theta \)
2. \( di - fg = -\sin \theta \cdot 1 - \sin \theta \cdot (-1) = -\sin \theta + \sin \theta = 0 \)
3. \( dh - eg = -\sin \theta \cdot (-\sin \theta) - 1 \cdot (-1) = \sin^2 \theta + 1 \)

Putting it all together:
\[ \Delta = 1 \cdot (1 + \sin^2 \theta) - \sin \theta \cdot 0 + 1 \cdot (\sin^2 \theta + 1) \]
This simplifies to:
\[ \Delta = 1 + \sin^2 \theta + \sin^2 \theta + 1 = 2 + 2\sin^2 \theta \]

### Step 4: Analyze the range of \( \Delta \)
Now we need to find the range of \( \Delta \):
\[ \Delta = 2 + 2\sin^2 \theta \]

Since \( \sin^2 \theta \) varies from 0 to 1 for \( 0 \leq \theta \leq 2\pi \):
- Minimum value of \( \Delta \) occurs when \( \sin^2 \theta = 0 \):
\[ \Delta_{\text{min}} = 2 + 2 \cdot 0 = 2 \]
- Maximum value of \( \Delta \) occurs when \( \sin^2 \theta = 1 \):
\[ \Delta_{\text{max}} = 2 + 2 \cdot 1 = 4 \]

### Conclusion
Thus, the value of \( \Delta \) lies in the interval:
\[ 2 \leq \Delta \leq 4 \]