If `f(theta)=|{:(1,tantheta,1),(-tantheta,1,tantheta),(-1,-tantheta,1):}|` then the set `{f(theta):0lt=thetalt(pi)/(2)}` is

Options

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

Answer

To solve the problem, we need to evaluate the determinant given by the function \( f(\theta) \) and find its range as \( \theta \) varies from \( 0 \) to \( \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Write the Determinant**:
The function is defined as:
\[ f(\theta) = \left| \begin{array}{ccc} 1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1 \end{array} \right| \]

2. **Expand the Determinant**:
We will expand this determinant using the first row:
\[ f(\theta) = 1 \cdot \left| \begin{array}{cc} 1 & \tan \theta \\ -\tan \theta & 1 \end{array} \right| - \tan \theta \cdot \left| \begin{array}{cc} -\tan \theta & \tan \theta \\ -1 & 1 \end{array} \right| + 1 \cdot \left| \begin{array}{cc} -\tan \theta & 1 \\ -1 & -\tan \theta \end{array} \right| \]

3. **Calculate Each Minor**:
- The first minor:
\[ \left| \begin{array}{cc} 1 & \tan \theta \\ -\tan \theta & 1 \end{array} \right| = 1 \cdot 1 - (-\tan \theta) \cdot \tan \theta = 1 + \tan^2 \theta \]
- The second minor:
\[ \left| \begin{array}{cc} -\tan \theta & \tan \theta \\ -1 & 1 \end{array} \right| = (-\tan \theta) \cdot 1 - \tan \theta \cdot (-1) = -\tan \theta + \tan \theta = 0 \]
- The third minor:
\[ \left| \begin{array}{cc} -\tan \theta & 1 \\ -1 & -\tan \theta \end{array} \right| = (-\tan \theta)(-\tan \theta) - 1(-1) = \tan^2 \theta + 1 \]

4. **Combine the Results**:
Putting it all together:
\[ f(\theta) = 1(1 + \tan^2 \theta) - \tan \theta(0) + 1(\tan^2 \theta + 1) = (1 + \tan^2 \theta) + (\tan^2 \theta + 1) \]
\[ = 2 + 2\tan^2 \theta \]

5. **Express in Terms of Secant**:
We know that \( 1 + \tan^2 \theta = \sec^2 \theta \), so:
\[ f(\theta) = 2(1 + \tan^2 \theta) = 2\sec^2 \theta \]

6. **Determine the Range**:
As \( \theta \) varies from \( 0 \) to \( \frac{\pi}{2} \):
- At \( \theta = 0 \), \( \sec(0) = 1 \) so \( f(0) = 2 \cdot 1^2 = 2 \).
- As \( \theta \) approaches \( \frac{\pi}{2} \), \( \sec(\theta) \) approaches infinity.

Therefore, the range of \( f(\theta) \) is:
\[ [2, \infty) \]

### Final Answer:
The set \( \{ f(\theta) : 0 \leq \theta < \frac{\pi}{2} \} \) is \( [2, \infty) \).