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 where \( \sin A + \sin B = a \) and \( \cos A + \cos B = b \), we need to find \( \cos(A + B) \).

### Step-by-Step Solution:

1. **Start with the given equations:**
\[ \sin A + \sin B = a \quad (1) \]
\[ \cos A + \cos B = b \quad (2) \]

2. **Square both equations:**
\[ (\sin A + \sin B)^2 = a^2 \implies \sin^2 A + \sin^2 B + 2 \sin A \sin B = a^2 \quad (3) \]
\[ (\cos A + \cos B)^2 = b^2 \implies \cos^2 A + \cos^2 B + 2 \cos A \cos B = b^2 \quad (4) \]

3. **Use the Pythagorean identity:**
From the Pythagorean identity, we know:
\[ \sin^2 A + \cos^2 A = 1 \quad (5) \]
\[ \sin^2 B + \cos^2 B = 1 \quad (6) \]

4. **Add equations (3) and (4):**
\[ a^2 + b^2 = (\sin^2 A + \sin^2 B) + (\cos^2 A + \cos^2 B) + 2(\sin A \sin B + \cos A \cos B) \]
Using (5) and (6):
\[ a^2 + b^2 = 1 + 1 + 2(\sin A \sin B + \cos A \cos B) \]
\[ a^2 + b^2 = 2 + 2(\sin A \sin B + \cos A \cos B) \]

5. **Rearranging gives us:**
\[ \sin A \sin B + \cos A \cos B = \frac{a^2 + b^2 - 2}{2} \quad (7) \]

6. **Use the cosine addition formula:**
The cosine of the sum of two angles is given by:
\[ \cos(A + B) = \cos A \cos B - \sin A \sin B \]
We can express this as:
\[ \cos(A + B) = \cos A \cos B - \sin A \sin B \]

7. **Substituting from (7):**
From (7), we can substitute:
\[ \cos A \cos B = \frac{b^2 - (a^2 + b^2 - 2)}{2} = \frac{2 + b^2 - a^2}{2} \]
\[ \sin A \sin B = \frac{a^2 + b^2 - 2}{2} \]
Thus:
\[ \cos(A + B) = \frac{b^2 - a^2}{2} \]

8. **Final expression for \( \cos(A + B) \):**
\[ \cos(A + B) = \frac{b^2 - a^2}{a^2 + b^2 + 1} \]

### Final Answer:
\[ \cos(A + B) = \frac{b^2 - a^2}{a^2 + b^2 + 1} \]