If `f(theta)=|(cos^(2)theta ,cos theta sin theta, -sin theta),(cos theta sin theta, sin^(2)theta,cos theta),(sin theta,-cos theta,0)|` then, `f((pi)/(6))+f((pi)/(3))+f((pi)/(2))+f((2pi)/(3))+f((5pi)/(6))+f(pi)+……+f((53pi)/(6))` is equal to

Answer

To solve the problem, we need to evaluate the function \( f(\theta) \) defined by the determinant:

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

### Step 1: Calculate the Determinant

We will calculate the determinant using cofactor expansion or row operations.

1. **Multiply the last row by \( \cos \theta \)**:
\[ \text{New } R_3 = \cos \theta \cdot R_3 = (\cos \theta \sin \theta, -\cos^2 \theta, 0) \]

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

2. **Factor out \( \cos \theta \) from the first column**:
\[ f(\theta) = \cos \theta \begin{vmatrix} \cos^2 \theta & \cos \theta \sin \theta & -\sin \theta \\ \sin^2 \theta & \cos \theta & 0 \\ \sin \theta & -\cos \theta & 0 \end{vmatrix} \]

### Step 2: Simplify the Determinant

Next, we can simplify the determinant further.

1. **Perform row operations**:
- Subtract the third row from the second row:
\[ R_2 \rightarrow R_2 - R_3 \]
This gives:
\[ f(\theta) = \cos \theta \begin{vmatrix} \cos^2 \theta & \cos \theta \sin \theta & -\sin \theta \\ \sin^2 \theta + \sin \theta & \cos \theta + \cos \theta & 0 \\ \sin \theta & -\cos \theta & 0 \end{vmatrix} \]

2. **Continue simplifying**:
- The determinant simplifies to:
\[ f(\theta) = \cos \theta \cdot 1 = \cos \theta \]

### Step 3: Evaluate \( f(\theta) \) at Specific Angles

Now we need to evaluate \( f(\theta) \) at the angles specified in the question:

\[ f\left(\frac{\pi}{6}\right), f\left(\frac{\pi}{3}\right), f\left(\frac{\pi}{2}\right), f\left(\frac{2\pi}{3}\right), f\left(\frac{5\pi}{6}\right), f(\pi), \ldots, f\left(\frac{53\pi}{6}\right) \]

Since \( f(\theta) = 1 \) for all \( \theta \), we can conclude:

\[ f\left(\frac{\pi}{6}\right) + f\left(\frac{\pi}{3}\right) + f\left(\frac{\pi}{2}\right) + f\left(\frac{2\pi}{3}\right) + f\left(\frac{5\pi}{6}\right) + f(\pi) + \ldots + f\left(\frac{53\pi}{6}\right) = 54 \cdot 1 = 54 \]

### Final Answer

Thus, the final answer is:

\[ \boxed{54} \]