Let `(sin(theta-alpha))/("sin"(theta-beta))=a/b a n d(cos(theta-alpha))/("cos"(theta-beta))=c/d t h e n(a c+b d)/(a d+b c)=`

Options

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

Explanation

To solve the given problem, we start with the equations provided and manipulate them step by step.

### Step 1: Set up the equations
We are given:
\[ \frac{\sin(\theta - \alpha)}{\sin(\theta - \beta)} = \frac{a}{b} \]
This implies:
\[ a \cdot \sin(\theta - \beta) = b \cdot \sin(\theta - \alpha) \tag{1} \]

Similarly, we have:
\[ \frac{\cos(\theta - \alpha)}{\cos(\theta - \beta)} = \frac{c}{d} \]
This implies:
\[ c \cdot \cos(\theta - \beta) = d \cdot \cos(\theta - \alpha) \tag{2} \]

### Step 2: Express \( a, b, c, d \) in terms of \( k_1, k_2 \)
From equations (1) and (2), we can express \( a \) and \( b \) in terms of a constant \( k_1 \) and \( c \) and \( d \) in terms of another constant \( k_2 \):
\[ \frac{a}{\sin(\theta - \alpha)} = k_1 \quad \text{and} \quad \frac{b}{\sin(\theta - \beta)} = k_1 \]
\[ \frac{c}{\cos(\theta - \alpha)} = k_2 \quad \text{and} \quad \frac{d}{\cos(\theta - \beta)} = k_2 \]

### Step 3: Substitute \( a, b, c, d \)
Now substituting \( a, b, c, d \) into the expression we need to evaluate:
\[ \frac{ac + bd}{ad + bc} \]
Substituting:
\[ = \frac{(k_1 \sin(\theta - \alpha))(k_2 \cos(\theta - \alpha)) + (k_1 \sin(\theta - \beta))(k_2 \cos(\theta - \beta))}{(k_1 \sin(\theta - \alpha))(k_2 \cos(\theta - \beta)) + (k_1 \sin(\theta - \beta))(k_2 \cos(\theta - \alpha))} \]

### Step 4: Factor out constants
Factoring out \( k_1 k_2 \) from both the numerator and denominator:
\[ = \frac{k_1 k_2 \left( \sin(\theta - \alpha) \cos(\theta - \alpha) + \sin(\theta - \beta) \cos(\theta - \beta) \right)}{k_1 k_2 \left( \sin(\theta - \alpha) \cos(\theta - \beta) + \sin(\theta - beta) \cos(\theta - alpha) \right)} \]
This simplifies to:
\[ = \frac{\sin(\theta - \alpha) \cos(\theta - \alpha) + \sin(\theta - \beta) \cos(\theta - \beta)}{\sin(\theta - \alpha) \cos(\theta - \beta) + \sin(\theta - \beta) \cos(\theta - \alpha)} \]

### Step 5: Apply trigonometric identities
Using the identity \( \sin A \cos B + \sin B \cos A = \sin(A + B) \):
- The numerator becomes:
\[ \frac{1}{2} \left( \sin(2(\theta - \alpha)) + \sin(2(\theta - \beta)) \right) \]
- The denominator becomes:
\[ \sin(\theta - \alpha + \theta - \beta) = \sin(2\theta - (\alpha + \beta)) \]

### Step 6: Final simplification
Thus, we have:
\[ \frac{\frac{1}{2} \left( \sin(2\theta - 2\alpha) + \sin(2\theta - 2\beta) \right)}{\sin(2\theta - \alpha - \beta)} \]
Using the sine addition formula:
\[ = \frac{\sin(2\theta - 2\alpha) + \sin(2\theta - 2\beta)}{2 \sin(2\theta - \alpha - \beta)} \]

### Step 7: Conclusion
After simplification, we find that:
\[ \frac{ac + bd}{ad + bc} = \cos(\alpha - \beta) \]

Thus, the final answer is:
\[ \cos(\alpha - \beta) \]