If `f(alpha,beta)=|(cos alpha,-sin alpha,1),(sin alpha,cos alpha,1),(cos(alpha+beta),-sin(alpha+beta),1)|,`then

Options

A. f(300,200=f(400,200)
B. f(200,400)=f(200,600)
C. f(100,200)=f(200,200)
D. none of these

Explanation

To solve the problem, we need to evaluate the determinant given by the function \( f(\alpha, \beta) \):

\[ f(\alpha, \beta) = \begin{vmatrix} \cos \alpha & -\sin \alpha & 1 \\ \sin \alpha & \cos \alpha & 1 \\ \cos(\alpha + \beta) & -\sin(\alpha + \beta) & 1 \end{vmatrix} \]

### Step 1: Expand the Determinant

We can expand the determinant using the method of cofactor expansion along the third column:

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

### Step 2: Calculate the 2x2 Determinants

1. The first determinant:

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

2. The second determinant:

\[ \begin{vmatrix} \cos \alpha & -\sin \alpha \\ \cos(\alpha + \beta) & -\sin(\alpha + \beta) \end{vmatrix} = -\cos \alpha \sin(\alpha + \beta) + \sin \alpha \cos(\alpha + \beta) \]
Using the sine addition formula, we have:

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

Thus, the determinant simplifies to:

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

3. The third determinant:

\[ \begin{vmatrix} \sin \alpha & \cos \alpha \\ \cos(\alpha + \beta) & -\sin(\alpha + \beta) \end{vmatrix} = -\sin \alpha \sin(\alpha + \beta) - \cos \alpha \cos(\alpha + \beta) \]
Using the cosine addition formula, we have:

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

Thus, the determinant simplifies to:

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

### Step 3: Combine the Results

Now, substituting back into our expression for \( f(\alpha, \beta) \):

\[ f(\alpha, \beta) = 1 - \sin \beta + \text{(third determinant)} \]

### Step 4: Final Expression

After simplifying, we find:

\[ f(\alpha, \beta) = 1 - \sin \beta + \text{(terms from the third determinant)} \]

### Conclusion

The final expression for \( f(\alpha, \beta) \) can be further simplified based on specific values of \( \alpha \) and \( \beta \).