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 of these

Answer

To find the value of \( f(x) \) given the determinant:

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

we will evaluate the determinant step by step.

### Step 1: Expand the Determinant

We can use the cofactor expansion along the first row:

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

### Step 2: Calculate the 2x2 Determinants

1. For the first determinant:

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

2. For the second determinant:

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

3. For the third determinant:

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

### Step 3: Substitute Back into the Expression for \( f(x) \)

Now substituting the values back into the expression for \( f(x) \):

\[ f(x) = \cos^2 x (\cos^2 x) - \cos x \sin x (-\cos x \sin x) - \sin x (0) \]

This simplifies to:

\[ f(x) = \cos^4 x + \cos^2 x \sin^2 x \]

### Step 4: Factor the Expression

We can factor the expression:

\[ f(x) = \cos^2 x (\cos^2 x + \sin^2 x) \]

Using the Pythagorean identity \( \cos^2 x + \sin^2 x = 1 \):

\[ f(x) = \cos^2 x \cdot 1 = \cos^2 x \]

### Step 5: Conclusion

Since \( \cos^2 x \) varies between 0 and 1 for all \( x \), we can conclude that \( f(x) \) is not constant for all \( x \) but rather depends on \( x \).

### Final Result

Thus, the final result is:

\[ f(x) = \cos^2 x \]

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