`|(sin alpha, cosalpha,sin(alpha+delta)),(sinbeta, cos beta,sin(beta+delta)),(singamma,cosgamma,sin(gamma+delta))|=`

Options

A. 0
B. 1
C. `1+sin alpha sin beta sin gamma`
D. `1-(sin alpha-sin beta)(sin beta-sin gamma)(sin gamma-sinalpha)`

Answer

To solve the determinant given by

\[ D = \begin{vmatrix} \sin \alpha & \cos \alpha & \sin(\alpha + \delta) \\ \sin \beta & \cos \beta & \sin(\beta + \delta) \\ \sin \gamma & \cos \gamma & \sin(\gamma + \delta) \end{vmatrix} \]

we will follow these steps:

### Step 1: Expand the determinant using the sine addition formula

The sine addition formula states that

\[ \sin(A + B) = \sin A \cos B + \cos A \sin B. \]

Using this formula, we can express \(\sin(\alpha + \delta)\), \(\sin(\beta + \delta)\), and \(\sin(\gamma + \delta)\):

\[ \sin(\alpha + \delta) = \sin \alpha \cos \delta + \cos \alpha \sin \delta, \]
\[ \sin(\beta + \delta) = \sin \beta \cos \delta + \cos \beta \sin \delta, \]
\[ \sin(\gamma + \delta) = \sin \gamma \cos \delta + \cos \gamma \sin \delta. \]

### Step 2: Substitute these expressions into the determinant

Now we can substitute these expressions into the determinant:

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

### Step 3: Apply column operations

We can simplify the determinant by performing column operations. Let's replace the third column \(C_3\) with \(C_3 - \sin \delta C_2 - \cos \delta C_1\):

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

### Step 4: Evaluate the determinant

Now we can see that the third column consists entirely of zeros. According to the properties of determinants, if any row or column is entirely zero, the value of the determinant is zero.

Thus, we conclude that:

\[ D = 0. \]

### Final Answer

The value of the determinant is

\[ \boxed{0}. \]