If `A = |(sin (theta + alpha),cos (theta + alpha),1),(sin (theta + beta),cos (theta + beta),1),(sin (theta + gamma),cos (theta + gamma),1)|`, then

Options

A. `A = 0 " for all " theta`
B. A is an odd function of `theta`
C. A = 0 for `theta = alpha + beta + gamma`
D. A is independent of `theta`

Explanation

To solve the determinant \( A = \begin{vmatrix} \sin(\theta + \alpha) & \cos(\theta + \alpha) & 1 \\ \sin(\theta + \beta) & \cos(\theta + \beta) & 1 \\ \sin(\theta + \gamma) & \cos(\theta + \gamma) & 1 \end{vmatrix} \), we will follow these steps:

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

\[ A = \begin{vmatrix} \sin(\theta + \alpha) & \cos(\theta + \alpha) & 1 \\ \sin(\theta + \beta) & \cos(\theta + \beta) & 1 \\ \sin(\theta + \gamma) & \cos(\theta + \gamma) & 1 \end{vmatrix} \]

### Step 2: Apply row operations
We will perform row operations to simplify the determinant. Specifically, we will subtract the first row from the second and third rows:

\[ R_2 \rightarrow R_2 - R_1 \]
\[ R_3 \rightarrow R_3 - R_1 \]

This gives us:

\[ A = \begin{vmatrix} \sin(\theta + \alpha) & \cos(\theta + \alpha) & 1 \\ \sin(\theta + \beta) - \sin(\theta + \alpha) & \cos(\theta + \beta) - \cos(\theta + \alpha) & 0 \\ \sin(\theta + \gamma) - \sin(\theta + \alpha) & \cos(\theta + \gamma) - \cos(\theta + \alpha) & 0 \end{vmatrix} \]

### Step 3: Simplify the determinant
Since the last column now consists of zeros except for the first row, we can expand the determinant along the third column:

\[ A = 1 \cdot \begin{vmatrix} \sin(\theta + \beta) - \sin(\theta + \alpha) & \cos(\theta + \beta) - \cos(\theta + \alpha) \\ \sin(\theta + \gamma) - \sin(\theta + \alpha) & \cos(\theta + \gamma) - \cos(\theta + \alpha) \end{vmatrix} \]

### Step 4: Use the sine and cosine subtraction formulas
Using the identities for sine and cosine differences, we can express the determinant in terms of sine and cosine:

\[ \sin(\theta + \beta) - \sin(\theta + \alpha) = 2 \sin\left(\frac{\beta - \alpha}{2}\right) \cos\left(\theta + \frac{\beta + \alpha}{2}\right) \]
\[ \cos(\theta + \beta) - \cos(\theta + \alpha) = -2 \sin\left(\frac{\beta - \alpha}{2}\right) \sin\left(\theta + \frac{\beta + \alpha}{2}\right) \]

### Step 5: Substitute back into the determinant
Substituting these back into the determinant gives:

\[ A = 2 \sin\left(\frac{\beta - \alpha}{2}\right) \cdot (-2 \sin\left(\frac{\beta - \alpha}{2}\right) \sin\left(\theta + \frac{\beta + \alpha}{2}\right)) = -4 \sin^2\left(\frac{\beta - \alpha}{2}\right) \sin\left(\theta + \frac{\beta + \alpha}{2}\right) \]

### Step 6: Analyze independence from theta
Notice that the expression contains terms that are independent of \(\theta\). Thus, we conclude that \(A\) is independent of \(\theta\).

### Final Result
The final expression shows that \(A\) does not depend on \(\theta\):

\[ A = 4 \sin\left(\frac{\beta - \alpha}{2}\right) \sin\left(\frac{\alpha - \gamma}{2}\right) \sin\left(\frac{\gamma - \beta}{2}\right) \]

### Conclusion
Thus, the correct option is that \(A\) is independent of \(\theta\).

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