The value of `f(pi/6)` where `f(theta)=|{:(cos^2theta,costhetasintheta,-sintheta),(costhetasintheta,sin^2theta,costheta),(sintheta,-costheta,0):}|`

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find the value of \( f\left(\frac{\pi}{6}\right) \) where

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

we will calculate the determinant step by step.

### Step 1: Write the determinant
We start with the determinant as given:

\[ D = \begin{vmatrix} \cos^2 \theta & \cos \theta \sin \theta & -\sin \theta \\ \cos \theta \sin \theta & \sin^2 \theta & \cos \theta \\ \sin \theta & -\cos \theta & 0 \end{vmatrix} \]

### Step 2: Expand the determinant
We can expand the determinant using the first row. The formula for the determinant of a 3x3 matrix is:

\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the remaining rows.

Using the first row:

\[ D = \cos^2 \theta \begin{vmatrix} \sin^2 \theta & \cos \theta \\ -\cos \theta & 0 \end{vmatrix} - \cos \theta \sin \theta \begin{vmatrix} \cos \theta \sin \theta & \cos \theta \\ \sin \theta & 0 \end{vmatrix} - \sin \theta \begin{vmatrix} \cos \theta \sin \theta & \sin^2 \theta \\ \sin \theta & -\cos \theta \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants
Now we calculate each of the 2x2 determinants:

1. For the first determinant:

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

2. For the second determinant:

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

3. For the third determinant:

\[ \begin{vmatrix} \cos \theta \sin \theta & \sin^2 \theta \\ \sin \theta & -\cos \theta \end{vmatrix} = (-\cos \theta \cdot \cos \theta \sin \theta) - (\sin^2 \theta \cdot \sin \theta) = -\cos^2 \theta \sin \theta - \sin^3 \theta \]

### Step 4: Substitute back into the determinant
Substituting these back into our expression for \( D \):

\[ D = \cos^2 \theta (\cos^2 \theta) - \cos \theta \sin \theta (-\cos \theta \sin \theta) - \sin \theta (-\cos^2 \theta \sin \theta - \sin^3 \theta) \]

This simplifies to:

\[ D = \cos^4 \theta + \cos^2 \theta \sin^2 \theta + \sin \theta (\cos^2 \theta \sin \theta + \sin^3 \theta) \]

### Step 5: Simplify further
Notice that \( \cos^2 \theta + \sin^2 \theta = 1 \):

\[ D = \cos^4 \theta + \cos^2 \theta \sin^2 \theta + \sin^2 \theta \cos^2 \theta + \sin^4 \theta \]

This can be factored as:

\[ D = (\cos^2 \theta + \sin^2 \theta)^2 = 1^2 = 1 \]

### Step 6: Conclusion
Thus, we find that:

\[ f(\theta) = 1 \]

Therefore,

\[ f\left(\frac{\pi}{6}\right) = 1 \]