If `Delta=|(sin alpha, cos alpha, sin alpha+cos beta),(sin beta, cos alpha, sin beta+cos beta),(sin gamma, cos alpha, sin gamma+cos beta)|` then `Delta` equals

Options

A. `sin alpha sin beta sin gamma`
B. `cos alpha sec beta tan gamma`
C. `sin alpha sin beta singamma +cos alpha cos beta cos gamma`
D. 0

Answer

To solve the determinant \( \Delta = \begin{vmatrix} \sin \alpha & \cos \alpha & \sin \alpha + \cos \beta \\ \sin \beta & \cos \alpha & \sin \beta + \cos \beta \\ \sin \gamma & \cos \alpha & \sin \gamma + \cos \beta \end{vmatrix} \), we will perform a series of steps to simplify it.

### Step 1: Apply Column Operations
We will modify the third column by subtracting the first column from it. This means we will replace \( C_3 \) with \( C_3 - C_1 \).

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

This simplifies to:

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

### Step 2: Factor Out Common Terms
Notice that the third column now has a common factor of \( \cos \beta \). We can factor this out:

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

### Step 3: Evaluate the Determinant
Now we need to evaluate the determinant:

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

We can see that the first two columns are dependent since they share the same second column. Thus, if two columns of a determinant are identical or proportional, the value of the determinant is zero.

### Conclusion
Thus, we conclude that:

\[ \Delta = 0 \]

### Final Answer
\(\Delta = 0\)

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