If f(x) = `|{:(cos^(2)x, cosx.sinx, -sin x),(cos x sinx ,sin^(2)x ,cos x),(sinx , -cos x , 0):}|` then for all x

Options

A. f(x) = 0
B. f(x) = 1
C. f(x) = 2
D. None ot these

Answer

To solve the given determinant problem, we need to evaluate the determinant \( f(x) \) defined as:

\[ f(x) = \begin{vmatrix} \cos^2 x & \cos x \sin x & -\sin x \\ \cos x \sin x & \sin^2 x & \cos x \\ \sin x & -\cos x & 0 \end{vmatrix} \]

### Step 1: Apply Column Operations
We will perform column operations to simplify the determinant.

1. Replace \( C_1 \) with \( C_1 - \sin x \cdot C_3 \).
2. Replace \( C_2 \) with \( C_2 - \cos x \cdot C_3 \).

After applying these operations, the determinant becomes:

\[ f(x) = \begin{vmatrix} 1 & 0 & \sin x \\ 0 & 1 & -\cos x \\ 0 & \sin x & \cos x \end{vmatrix} \]

### Step 2: Simplify the Determinant
Now, we can simplify the determinant further. The determinant can be expanded along the first column:

\[ f(x) = 1 \cdot \begin{vmatrix} 1 & -\cos x \\ \sin x & \cos x \end{vmatrix} \]

### Step 3: Calculate the 2x2 Determinant
Now, we calculate the 2x2 determinant:

\[ \begin{vmatrix} 1 & -\cos x \\ \sin x & \cos x \end{vmatrix} = (1)(\cos x) - (-\cos x)(\sin x) = \cos x + \sin x \cos x \]

### Step 4: Substitute Back
Thus, we have:

\[ f(x) = 1 \cdot (\cos x + \sin x \cos x) = \cos x + \sin x \cos x \]

### Step 5: Factor Out Common Terms
We can factor out \( \cos x \):

\[ f(x) = \cos x (1 + \sin x) \]

### Step 6: Use Pythagorean Identity
Now, we apply the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \). We can also evaluate the determinant directly by recognizing that:

\[ \sin^2 x + \cos^2 x = 1 \]

### Conclusion
Thus, we conclude that:

\[ f(x) = 1 \quad \text{for all } x \]