If the lines `x(sinalpha+sinbeta)-ysin(alpha-beta)=3` and `x(cosalpha+cosbeta)+ycos(alpha-beta)=5` are perpendicular then `sin2alpha+sin2beta` is equal to:

Options

A. `sin(alpha-beta)-2sin(alpha+beta)`
B. `sin2(alpha-beta)-2sin(alpha+beta)`
C. `2sin(alpha-beta)-sin(alpha+beta)`
D. `sin2(alpha-beta)-sin(alpha+beta)`

Answer

To solve the problem, we need to determine the value of \( \sin 2\alpha + \sin 2\beta \) given that the lines represented by the equations are perpendicular.

### Step 1: Write down the equations
The equations given are:
1. \( x(\sin \alpha + \sin \beta) - y \sin(\alpha - \beta) = 3 \) (Equation 1)
2. \( x(\cos \alpha + \cos \beta) + y \cos(\alpha - \beta) = 5 \) (Equation 2)

### Step 2: Identify the slopes of the lines
The slope of a line in the form \( Ax + By + C = 0 \) is given by \( m = -\frac{A}{B} \).

For Equation 1:
- \( A = \sin \alpha + \sin \beta \)
- \( B = -\sin(\alpha - \beta) \)

Thus, the slope \( m_1 \) is:
\[ m_1 = -\frac{\sin \alpha + \sin \beta}{-\sin(\alpha - \beta)} = \frac{\sin \alpha + \sin \beta}{\sin(\alpha - \beta)} \]

For Equation 2:
- \( A = \cos \alpha + \cos \beta \)
- \( B = \cos(\alpha - \beta) \)

Thus, the slope \( m_2 \) is:
\[ m_2 = -\frac{\cos \alpha + \cos \beta}{\cos(\alpha - \beta)} \]

### Step 3: Use the condition for perpendicular lines
Two lines are perpendicular if the product of their slopes is -1:
\[ m_1 \cdot m_2 = -1 \]

Substituting the slopes:
\[ \left(\frac{\sin \alpha + \sin \beta}{\sin(\alpha - \beta)}\right) \cdot \left(-\frac{\cos \alpha + \cos \beta}{\cos(\alpha - \beta)}\right) = -1 \]

This simplifies to:
\[ -\frac{(\sin \alpha + \sin \beta)(\cos \alpha + \cos \beta)}{\sin(\alpha - \beta) \cos(\alpha - \beta)} = -1 \]

Removing the negative sign gives:
\[ \frac{(\sin \alpha + \sin \beta)(\cos \alpha + \cos \beta)}{\sin(\alpha - \beta) \cos(\alpha - \beta)} = 1 \]

### Step 4: Cross-multiply to simplify
Cross-multiplying gives:
\[ (\sin \alpha + \sin \beta)(\cos \alpha + \cos \beta) = \sin(\alpha - \beta) \cos(\alpha - \beta) \]

### Step 5: Use trigonometric identities
Using the identity \( \sin A \cos A = \frac{1}{2} \sin(2A) \):
\[ \sin(\alpha - \beta) \cos(\alpha - \beta) = \frac{1}{2} \sin(2(\alpha - \beta)) \]

### Step 6: Expand the left-hand side
Expanding the left-hand side:
\[ \sin \alpha \cos \alpha + \sin \alpha \cos \beta + \sin \beta \cos \alpha + \sin \beta \cos \beta \]

Using the identity \( \sin 2\alpha = 2 \sin \alpha \cos \alpha \) and \( \sin 2\beta = 2 \sin \beta \cos \beta \):
\[ \frac{1}{2} \sin 2\alpha + \frac{1}{2} \sin 2\beta + \sin \alpha \cos \beta + \sin \beta \cos \alpha = \frac{1}{2} \sin(2(\alpha - \beta)) \]

### Step 7: Solve for \( \sin 2\alpha + \sin 2\beta \)
From the equation, we can isolate \( \sin 2\alpha + \sin 2\beta \):
\[ \sin 2\alpha + \sin 2\beta = \sin(2(\alpha - \beta)) - 2(\sin \alpha \cos \beta + \sin \beta \cos \alpha) \]

Using the identity \( \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \), we can express it in terms of \( \alpha \) and \( \beta \).

### Final Answer
Thus, the final value of \( \sin 2\alpha + \sin 2\beta \) can be expressed as:
\[ \sin 2\alpha + \sin 2\beta = \sin(2(\alpha - \beta)) - 2 \sin(\alpha + \beta) \]