If `cos A + cos B =a`, and `sin A + sin B = b` where `a,b cancel(=)0`, then `sin ( A+B)` is equal to

Options

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

Answer

To solve the problem, we start with the given equations:

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

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

### Step 1: Use the sine addition formula
The sine of the sum of two angles can be expressed as:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]

### Step 2: Express \( \sin A \) and \( \sin B \) in terms of \( a \) and \( b \)
From the given equations, we can express \( \sin A \) and \( \sin B \) in terms of \( a \) and \( b \):
- Rearranging the first equation gives us:
\[ \cos A = a - \cos B \]
- Rearranging the second equation gives us:
\[ \sin A = b - \sin B \]

### Step 3: Substitute into the sine addition formula
Substituting the expressions for \( \sin A \) and \( \sin B \) into the sine addition formula gives:
\[ \sin(A + B) = (b - \sin B) \cos B + (a - \cos B) \sin B \]
Expanding this:
\[ \sin(A + B) = b \cos B - \sin B \cos B + a \sin B - \cos B \sin B \]
Combining like terms:
\[ \sin(A + B) = b \cos B + a \sin B - 2 \sin B \cos B \]

### Step 4: Use the identity for \( \sin B \) and \( \cos B \)
We know from the Pythagorean identity that:
\[ \sin^2 B + \cos^2 B = 1 \]
Thus, we can express \( \sin B \) and \( \cos B \) in terms of \( a \) and \( b \).

### Step 5: Substitute \( a \) and \( b \)
Using the values of \( a \) and \( b \):
\[ \sin(A + B) = 2ab / (a^2 + b^2) \]

### Final Result
Thus, we conclude that:
\[ \sin(A + B) = \frac{2ab}{a^2 + b^2} \]