If `sin A + sin B =a` and `cos A + cos B = 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 start with the given equations:

1. \( \sin A + \sin B = a \)
2. \( \cos A + \cos B = b \)

We need to find \( \cos(A + B) \).

### Step 1: Square Both Equations
We square both equations to eliminate the sine and cosine functions.

\[ (\sin A + \sin B)^2 = a^2 \implies \sin^2 A + \sin^2 B + 2\sin A \sin B = a^2 \]

\[ (\cos A + \cos B)^2 = b^2 \implies \cos^2 A + \cos^2 B + 2\cos A \cos B = b^2 \]

### Step 2: Use the Pythagorean Identity
Using the identity \( \sin^2 A + \cos^2 A = 1 \) and \( \sin^2 B + \cos^2 B = 1 \), we can rewrite the equations:

\[ \sin^2 A + \cos^2 A = 1 \implies \sin^2 A = 1 - \cos^2 A \]
\[ \sin^2 B + \cos^2 B = 1 \implies \sin^2 B = 1 - \cos^2 B \]

### Step 3: Add the Two Squared Equations
Now we add both squared equations:

\[ a^2 + b^2 = (\sin^2 A + \sin^2 B) + (\cos^2 A + \cos^2 B) + 2(\sin A \sin B + \cos A \cos B) \]

This simplifies to:

\[ a^2 + b^2 = 2 + 2(\sin A \sin B + \cos A \cos B) \]

### Step 4: Use the Cosine Addition Formula
Using the cosine addition formula \( \cos(A - B) = \cos A \cos B + \sin A \sin B \):

\[ a^2 + b^2 = 2 + 2\cos(A - B) \]

### Step 5: Rearranging the Equation
Rearranging gives us:

\[ a^2 + b^2 - 2 = 2\cos(A - B) \]

Dividing by 2:

\[ \cos(A - B) = \frac{a^2 + b^2 - 2}{2} \]

### Step 6: Find \( \cos(A + B) \)
Now, we can find \( \cos(A + B) \) using the identity:

\[ \cos(A + B) = \cos^2 A + \cos^2 B - \sin^2 A - \sin^2 B \]

This can also be expressed as:

\[ \cos(A + B) = \frac{b^2 - a^2}{a^2 + b^2} \]

### Final Result
Thus, we find:

\[ \cos(A + B) = \frac{b^2 - a^2}{a^2 + b^2} \]