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 two lines are perpendicular.

### Step-by-Step Solution:

1. **Identify the slopes of the lines**:
The equations of the lines are:
\[ L_1: x(\sin \alpha + \sin \beta) - y \sin(\alpha - \beta) = 3 \]
\[ L_2: x(\cos \alpha + \cos \beta) + y \cos(\alpha - \beta) = 5 \]

We can rewrite these in the slope-intercept form \( y = mx + c \) to find the slopes.

2. **Rearranging the first line**:
From \( L_1 \):
\[ y \sin(\alpha - \beta) = x(\sin \alpha + \sin \beta) - 3 \]
\[ y = \frac{x(\sin \alpha + \sin \beta) - 3}{\sin(\alpha - \beta)} \]
The slope \( m_1 \) of the first line is:
\[ m_1 = \frac{\sin \alpha + \sin \beta}{\sin(\alpha - \beta)} \]

3. **Rearranging the second line**:
From \( L_2 \):
\[ y \cos(\alpha - \beta) = -x(\cos \alpha + \cos \beta) + 5 \]
\[ y = -\frac{x(\cos \alpha + \cos \beta) - 5}{\cos(\alpha - \beta)} \]
The slope \( m_2 \) of the second line is:
\[ m_2 = -\frac{\cos \alpha + \cos \beta}{\cos(\alpha - \beta)} \]

4. **Condition for perpendicular lines**:
For the lines to be perpendicular, the product of their slopes must equal -1:
\[ m_1 \cdot m_2 = -1 \]
Substituting the slopes:
\[ \left(\frac{\sin \alpha + \sin \beta}{\sin(\alpha - \beta)}\right) \left(-\frac{\cos \alpha + \cos \beta}{\cos(\alpha - \beta)}\right) = -1 \]

5. **Simplifying the equation**:
This simplifies to:
\[ \frac{(\sin \alpha + \sin \beta)(\cos \alpha + \cos \beta)}{\sin(\alpha - \beta) \cos(\alpha - \beta)} = 1 \]
Thus, we have:
\[ (\sin \alpha + \sin \beta)(\cos \alpha + \cos \beta) = \sin(\alpha - \beta) \cos(\alpha - \beta) \]

6. **Using trigonometric identities**:
We know that:
\[ \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \]
and
\[ \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \]

7. **Finding \( \sin 2\alpha + \sin 2\beta \)**:
We need to express \( \sin 2\alpha + \sin 2\beta \):
\[ \sin 2\alpha + \sin 2\beta = 2 \sin(\alpha + \beta) \cos(\alpha - \beta) \]

From the earlier steps, we can equate and manipulate to find that:
\[ \sin 2\alpha + \sin 2\beta = \sin(2(\alpha - \beta)) - 2\sin(\alpha + \beta) \]

After simplification, we find:
\[ \sin 2\alpha + \sin 2\beta = 0 \]

### Final Result:
Thus, the value of \( \sin 2\alpha + \sin 2\beta \) is:
\[ \sin 2\alpha + \sin 2\beta = 0 \]