If `f(x)=|{:(sectheta," "tan^(2)theta,1),(thetasecx," "tanx,x),(" "1,tanx-tantheta,0):}|`, then `f'(theta)` is

Options

A. 0
B. -1
C. independent of `theta`
D. none of these

Explanation

To find \( f'(\theta) \) for the function defined by the determinant

\[ f(\theta) = \begin{vmatrix} \sec \theta & \tan^2 \theta & 1 \\ \theta & \sec x & \tan x \\ x & 1 & \tan x - \tan \theta \end{vmatrix} \]

we will differentiate this determinant with respect to \( \theta \).

### Step-by-Step Solution

1. **Write down the determinant**:
\[ f(\theta) = \begin{vmatrix} \sec \theta & \tan^2 \theta & 1 \\ \theta & \sec x & \tan x \\ x & 1 & \tan x - \tan \theta \end{vmatrix} \]

2. **Differentiate with respect to \( \theta \)**:
To differentiate a determinant, we can use the property that the derivative of a determinant can be computed by differentiating each row with respect to the variable of differentiation (here, \( \theta \)).

3. **Differentiate the first row**:
- The derivative of \( \sec \theta \) is \( \sec \theta \tan \theta \).
- The derivative of \( \tan^2 \theta \) is \( 2 \tan \theta \sec^2 \theta \).
- The derivative of \( 1 \) is \( 0 \).

Thus, the first row becomes:
\[ \begin{vmatrix} \sec \theta \tan \theta & 2 \tan \theta \sec^2 \theta & 0 \\ \theta & \sec x & \tan x \\ x & 1 & \tan x - \tan \theta \end{vmatrix} \]

4. **Differentiate the second row**:
The second row does not depend on \( \theta \), so it remains unchanged:
\[ \begin{vmatrix} \sec \theta \tan \theta & 2 \tan \theta \sec^2 \theta & 0 \\ \theta & \sec x & \tan x \\ x & 1 & \tan x - \tan \theta \end{vmatrix} \]

5. **Differentiate the third row**:
- The derivative of \( \tan x - \tan \theta \) with respect to \( \theta \) is \( -\sec^2 \theta \).

Thus, the third row becomes:
\[ \begin{vmatrix} \sec \theta \tan \theta & 2 \tan \theta \sec^2 \theta & 0 \\ \theta & \sec x & \tan x \\ x & 1 & -\sec^2 \theta \end{vmatrix} \]

6. **Evaluate the determinant**:
Now we can evaluate the determinant using the cofactor expansion along the first row.

The determinant can be calculated as:
\[ f'(\theta) = \sec \theta \tan \theta \cdot \begin{vmatrix} \sec x & \tan x \\ 1 & -\sec^2 \theta \end{vmatrix} + 2 \tan \theta \sec^2 \theta \cdot \begin{vmatrix} \theta & \tan x \\ x & -\sec^2 \theta \end{vmatrix} \]

7. **Simplify the determinants**:
Calculate the 2x2 determinants:
- For the first determinant:
\[ \begin{vmatrix} \sec x & \tan x \\ 1 & -\sec^2 \theta \end{vmatrix} = -\sec^2 \theta \sec x - \tan x \]
- For the second determinant:
\[ \begin{vmatrix} \theta & \tan x \\ x & -\sec^2 \theta \end{vmatrix} = -\theta \sec^2 \theta - x \tan x \]

8. **Put it all together**:
Substitute back into the expression for \( f'(\theta) \):
\[ f'(\theta) = \sec \theta \tan \theta (-\sec^2 \theta \sec x - \tan x) + 2 \tan \theta \sec^2 \theta (-\theta \sec^2 \theta - x \tan x) \]

9. **Final expression**:
After simplification, we find that:
\[ f'(\theta) = 1 \tan^2 \theta - \sec^2 \theta \]