If `alpha,beta and gamma` are such that `alpha+beta+gamma=0`, then `|(1,cos gamma,cosbeta),(cosgamma,1,cos alpha),(cosbeta,cos alpha,1)|`

Options

A. `cos alpha cos beta cos gamma`
B. `cos alpha + cos beta + cos gamma`
C. 1
D. none of these

Explanation

To solve the determinant given the condition \( \alpha + \beta + \gamma = 0 \), we will follow these steps:

### Step 1: Write the Determinant
We need to evaluate the determinant:

\[ D = \begin{vmatrix} 1 & \cos \gamma & \cos \beta \\ \cos \gamma & 1 & \cos \alpha \\ \cos \beta & \cos \alpha & 1 \end{vmatrix} \]

### Step 2: Expand the Determinant
We can expand the determinant using the first row:

\[ D = 1 \cdot \begin{vmatrix} 1 & \cos \alpha \\ \cos \alpha & 1 \end{vmatrix} - \cos \gamma \cdot \begin{vmatrix} \cos \gamma & \cos \alpha \\ \cos \beta & 1 \end{vmatrix} + \cos \beta \cdot \begin{vmatrix} \cos \gamma & 1 \\ \cos \beta & \cos \alpha \end{vmatrix} \]

### Step 3: Calculate the 2x2 Determinants
1. The first 2x2 determinant:

\[ \begin{vmatrix} 1 & \cos \alpha \\ \cos \alpha & 1 \end{vmatrix} = 1 \cdot 1 - \cos^2 \alpha = 1 - \cos^2 \alpha = \sin^2 \alpha \]

2. The second 2x2 determinant:

\[ \begin{vmatrix} \cos \gamma & \cos \alpha \\ \cos \beta & 1 \end{vmatrix} = \cos \gamma \cdot 1 - \cos \alpha \cdot \cos \beta = \cos \gamma - \cos \alpha \cos \beta \]

3. The third 2x2 determinant:

\[ \begin{vmatrix} \cos \gamma & 1 \\ \cos \beta & \cos \alpha \end{vmatrix} = \cos \gamma \cdot \cos \alpha - 1 \cdot \cos \beta = \cos \gamma \cos \alpha - \cos \beta \]

### Step 4: Substitute Back into the Determinant
Now substituting these back into the determinant expression:

\[ D = 1 \cdot \sin^2 \alpha - \cos \gamma (\cos \gamma - \cos \alpha \cos \beta) + \cos \beta (\cos \gamma \cos \alpha - \cos \beta) \]

### Step 5: Simplify the Expression
Expanding this gives:

\[ D = \sin^2 \alpha - \cos^2 \gamma + \cos \gamma \cos \alpha \cos \beta + \cos \beta \cos \gamma \cos \alpha - \cos^2 \beta \]

Combining like terms:

\[ D = \sin^2 \alpha - \cos^2 \gamma - \cos^2 \beta + 2 \cos \gamma \cos \alpha \cos \beta \]

### Step 6: Use the Condition \( \alpha + \beta + \gamma = 0 \)
From the condition \( \alpha + \beta + \gamma = 0 \), we have:

\[ \gamma = -(\alpha + \beta) \]

Using the cosine of a sum:

\[ \cos(\alpha + \beta) = \cos \gamma = \cos(-\gamma) = \cos \gamma \]

### Step 7: Final Simplification
Using the identity \( \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \):

Substituting this back into the determinant, we will find that the determinant simplifies to zero.

### Conclusion
Thus, the value of the determinant is:

\[ D = 0 \]