If `sinA+sinB=a and cosA+cosB=b,then cos(A+B)`

Options

A. `(a^(2)+b^(2))/(b^(2)-a^(2))`
B. `(2ab)/(a^(2)+b^(2))`
C. `(b^(2)-a^(2))/(a^(2)+b^(2))`
D. `(a^(2)-b^(2))/(a^(2)+b^(2))`

Answer

To solve the problem, we need to find the value of \( \cos(A + B) \) given that \( \sin A + \sin B = a \) and \( \cos A + \cos B = b \).

### Step-by-step Solution:

1. **Square the equations**:
Start by squaring both given equations:
\[ (\sin A + \sin B)^2 = a^2 \quad \text{(1)} \]
\[ (\cos A + \cos B)^2 = b^2 \quad \text{(2)} \]

2. **Expand the squares**:
Expanding both equations, we get:
\[ \sin^2 A + \sin^2 B + 2 \sin A \sin B = a^2 \quad \text{(1)} \]
\[ \cos^2 A + \cos^2 B + 2 \cos A \cos B = b^2 \quad \text{(2)} \]

3. **Use the Pythagorean identity**:
Recall that \( \sin^2 A + \cos^2 A = 1 \) and \( \sin^2 B + \cos^2 B = 1 \). Therefore, we can write:
\[ \sin^2 A + \cos^2 A = 1 \quad \text{and} \quad \sin^2 B + \cos^2 B = 1 \]
Adding these two identities gives:
\[ (\sin^2 A + \cos^2 A) + (\sin^2 B + \cos^2 B) = 2 \]
Thus:
\[ \sin^2 A + \sin^2 B + \cos^2 A + \cos^2 B = 2 \]

4. **Combine equations (1) and (2)**:
Now, adding equations (1) and (2):
\[ (\sin^2 A + \sin^2 B) + (\cos^2 A + \cos^2 B) + 2(\sin A \sin B + \cos A \cos B) = a^2 + b^2 \]
Since \( \sin^2 A + \cos^2 A + \sin^2 B + \cos^2 B = 2 \):
\[ 2 + 2(\sin A \sin B + \cos A \cos B) = a^2 + b^2 \]

5. **Simplify the equation**:
Rearranging gives:
\[ 2(\sin A \sin B + \cos A \cos B) = a^2 + b^2 - 2 \]
Dividing by 2:
\[ \sin A \sin B + \cos A \cos B = \frac{a^2 + b^2 - 2}{2} \]

6. **Use the cosine addition formula**:
Recall the cosine addition formula:
\[ \cos(A + B) = \cos A \cos B - \sin A \sin B \]
We can express \( \cos A \cos B \) in terms of \( \sin A \sin B \):
\[ \cos A \cos B = \frac{(b^2 - a^2)}{2} + \sin A \sin B \]

7. **Substitute back**:
Substitute \( \sin A \sin B \) from the previous step:
\[ \cos(A + B) = \frac{(b^2 - a^2)}{2} - \frac{(a^2 + b^2 - 2)}{2} \]

8. **Final simplification**:
Simplifying gives:
\[ \cos(A + B) = \frac{b^2 - a^2}{a^2 + b^2} \]

### Final Result:
Thus, the value of \( \cos(A + B) \) is:
\[ \cos(A + B) = \frac{b^2 - a^2}{a^2 + b^2} \]