`|[1,cos(alpha-beta), cos alpha] , [cos(alpha-beta),1,cos beta] , [cos alpha, cos beta, 1]|`

Options

A. `sin (alpha + beta)`
B. `sin alpha sin beta`
C. `1 + cos (alpha + beta)`
D. none of these

Answer

To find the value of the determinant

\[ D = \begin{vmatrix} 1 & \cos(\alpha - \beta) & \cos \alpha \\ \cos(\alpha - \beta) & 1 & \cos \beta \\ \cos \alpha & \cos \beta & 1 \end{vmatrix} \]

we will use properties of determinants and trigonometric identities.

### Step 1: Rewrite \(\cos(\alpha - \beta)\)

Using the cosine subtraction formula, we can express \(\cos(\alpha - \beta)\) as:

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

Now, substituting this into the determinant gives us:

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

### Step 2: Apply Column Operations

We will perform column operations to simplify the determinant. Let's modify column 1 and column 2:

- Replace column 1 with \(C_1 - \cos \alpha C_3\)
- Replace column 2 with \(C_2 - \cos \beta C_3\)

This gives us:

\[ D = \begin{vmatrix} 1 - \cos^2 \alpha & \sin \alpha \sin \beta & \cos \alpha \\ \sin \alpha \sin \beta & 1 - \cos^2 \beta & \cos \beta \\ 0 & 0 & 1 \end{vmatrix} \]

### Step 3: Simplify the Determinant

Using the identity \(1 - \cos^2 \theta = \sin^2 \theta\), we can rewrite the determinant as:

\[ D = \begin{vmatrix} \sin^2 \alpha & \sin \alpha \sin \beta & \cos \alpha \\ \sin \alpha \sin \beta & \sin^2 \beta & \cos \beta \\ 0 & 0 & 1 \end{vmatrix} \]

### Step 4: Expand the Determinant

We can expand the determinant along the third row:

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

Calculating this 2x2 determinant:

\[ = \sin^2 \alpha \cdot \sin^2 \beta - (\sin \alpha \sin \beta)(\sin \alpha \sin \beta) = \sin^2 \alpha \sin^2 \beta - \sin^2 \alpha \sin^2 \beta = 0 \]

### Conclusion

Thus, the value of the determinant \(D\) is:

\[ D = 0 \]