If `f(theta) = |(1,tan theta,1),(- tan theta,1,tan theta),(-1,-tan theta,1)|, " then the set " {f(theta ) : 0 le theta le (pi)/(2)}` is

Options

A. `(-oo,0] uu[2,oo)`
B. `[2,oo)`
C. `(-oo, 0) uu(0,oo)`
D. `(-oo, -1] uu [1, oo)`

Explanation

To solve the problem, we need to evaluate the determinant given by the function \( f(\theta) = \begin{vmatrix} 1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1 \end{vmatrix} \) and find the range of this function for \( 0 \leq \theta \leq \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Calculate the Determinant**:
We will use the formula for the determinant of a 3x3 matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix:
\[ A = \begin{pmatrix} 1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1 \end{pmatrix} \]
We label the elements as follows:
- \( a = 1, b = \tan \theta, c = 1 \)
- \( d = -\tan \theta, e = 1, f = \tan \theta \)
- \( g = -1, h = -\tan \theta, i = 1 \)

Now, substituting into the determinant formula:
\[ f(\theta) = 1 \cdot (1 \cdot 1 - \tan \theta \cdot (-\tan \theta)) - \tan \theta \cdot (-\tan \theta \cdot 1 - \tan \theta \cdot (-1)) + 1 \cdot (-\tan \theta \cdot (-\tan \theta) - 1 \cdot 1) \]

2. **Simplify Each Term**:
- The first term simplifies to:
\[ 1 \cdot (1 + \tan^2 \theta) = 1 + \tan^2 \theta \]
- The second term simplifies to:
\[ -\tan \theta \cdot (-\tan \theta - \tan \theta) = -\tan \theta \cdot (-2\tan \theta) = 2\tan^2 \theta \]
- The third term simplifies to:
\[ 1 \cdot (\tan^2 \theta - 1) = \tan^2 \theta - 1 \]

3. **Combine the Terms**:
Now, we combine all the terms:
\[ f(\theta) = (1 + \tan^2 \theta) + 2\tan^2 \theta + (\tan^2 \theta - 1) \]
Simplifying this gives:
\[ f(\theta) = 1 + \tan^2 \theta + 2\tan^2 \theta + \tan^2 \theta - 1 = 4\tan^2 \theta \]

4. **Determine the Range of \( f(\theta) \)**:
We know that \( \tan^2 \theta \) varies from \( 0 \) to \( \infty \) as \( \theta \) varies from \( 0 \) to \( \frac{\pi}{2} \). Therefore:
\[ f(\theta) = 4\tan^2 \theta \text{ varies from } 0 \text{ to } \infty. \]
Thus, the range of \( f(\theta) \) is:
\[ [0, \infty) \]

5. **Final Result**:
The set \( \{ f(\theta) : 0 \leq \theta \leq \frac{\pi}{2} \} \) is \( [0, \infty) \).