If `f(x)=|{:(cos(x+alpha),cos(x+beta),cos(x+gamma)),(sin(x+alpha),sin(x+beta),sin(x+gamma)),(sin(beta-gamma),sin(gamma-alpha),sin(alpha-beta)):}| and f(2) =5 "the" overset(20)underset(r=1)sumf(r)` is equal to

Options

A. 5
B. 10
C. 100
D. none of these

Answer

To solve the problem, we need to evaluate the function \( f(x) \) defined as the determinant of a 3x3 matrix involving trigonometric functions, and then compute the sum of this function from \( r = 1 \) to \( r = 20 \).

### Step-by-Step Solution:

1. **Understanding the Function**:
The function is given as:
\[ f(x) = \begin{vmatrix} \cos(x+\alpha) & \cos(x+\beta) & \cos(x+\gamma) \\ \sin(x+\alpha) & \sin(x+\beta) & \sin(x+\gamma) \\ \sin(\beta-\gamma) & \sin(\gamma-\alpha) & \sin(\alpha-\beta) \end{vmatrix} \]

2. **Evaluating the Determinant**:
We will expand the determinant. The first row will be expanded:
\[ f(x) = \sin(\beta - \gamma) \begin{vmatrix} \cos(x+\beta) & \cos(x+\gamma) \\ \sin(x+\beta) & \sin(x+\gamma) \end{vmatrix} - \sin(\gamma - \alpha) \begin{vmatrix} \cos(x+\alpha) & \cos(x+\gamma) \\ \sin(x+\alpha) & \sin(x+\gamma) \end{vmatrix} + \sin(\alpha - \beta) \begin{vmatrix} \cos(x+\alpha) & \cos(x+\beta) \\ \sin(x+\alpha) & \sin(x+\beta) \end{vmatrix} \]

3. **Using the Determinant Property**:
The determinant of a 2x2 matrix can be computed as:
\[ \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \]
Applying this to the 2x2 matrices in our expansion will yield sine differences.

4. **Simplifying the Expression**:
After expanding and simplifying, we find that:
\[ f(x) = C \]
where \( C \) is a constant that does not depend on \( x \).

5. **Given Value**:
We are given that \( f(2) = 5 \). Since \( f(x) \) is constant, we have:
\[ f(x) = 5 \quad \forall x \]

6. **Summation**:
We need to compute:
\[ \sum_{r=1}^{20} f(r) = f(1) + f(2) + f(3) + \ldots + f(20) \]
Since \( f(r) = 5 \) for all \( r \):
\[ \sum_{r=1}^{20} f(r) = 5 + 5 + 5 + \ldots + 5 \quad (\text{20 times}) \]
This results in:
\[ = 20 \times 5 = 100 \]

### Final Answer:
\[ \sum_{r=1}^{20} f(r) = 100 \]