If `sin alpha+sin beta=a` and `cos alpha-cos beta=b` then `tan((alpha-beta)/2)` is equal to-

Options

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

Answer

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

1. \( \sin \alpha + \sin \beta = a \)
2. \( \cos \alpha - \cos \beta = b \)

We need to find \( \tan\left(\frac{\alpha - \beta}{2}\right) \).

### Step 1: Use the sum-to-product identities

Using the sum-to-product identities, we can express \( \sin \alpha + \sin \beta \) and \( \cos \alpha - \cos \beta \):

- The formula for \( \sin \alpha + \sin \beta \) is:
\[ \sin \alpha + \sin \beta = 2 \sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) \]
Therefore, we have:
\[ 2 \sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) = a \]

- The formula for \( \cos \alpha - \cos \beta \) is:
\[ \cos \alpha - \cos \beta = -2 \sin\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right) \]
Therefore, we have:
\[ -2 \sin\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right) = b \]

### Step 2: Rearranging the equations

From the first equation, we can express \( \cos\left(\frac{\alpha - \beta}{2}\right) \):
\[ \cos\left(\frac{\alpha - \beta}{2}\right) = \frac{a}{2 \sin\left(\frac{\alpha + \beta}{2}\right)} \]

From the second equation, we can express \( \sin\left(\frac{\alpha - \beta}{2}\right) \):
\[ \sin\left(\frac{\alpha - \beta}{2}\right) = -\frac{b}{2 \sin\left(\frac{\alpha + \beta}{2}\right)} \]

### Step 3: Finding \( \tan\left(\frac{\alpha - \beta}{2}\right) \)

Now, we can find \( \tan\left(\frac{\alpha - \beta}{2}\right) \) using the definition of tangent:
\[ \tan\left(\frac{\alpha - \beta}{2}\right) = \frac{\sin\left(\frac{\alpha - \beta}{2}\right)}{\cos\left(\frac{\alpha - \beta}{2}\right)} \]

Substituting the expressions we found:
\[ \tan\left(\frac{\alpha - \beta}{2}\right) = \frac{-\frac{b}{2 \sin\left(\frac{\alpha + \beta}{2}\right)}}{\frac{a}{2 \sin\left(\frac{\alpha + \beta}{2}\right)}} \]

### Step 4: Simplifying the expression

The \( 2 \sin\left(\frac{\alpha + \beta}{2}\right) \) cancels out:
\[ \tan\left(\frac{\alpha - \beta}{2}\right) = \frac{-b}{a} \]

### Final Result

Thus, we conclude that:
\[ \tan\left(\frac{\alpha - \beta}{2}\right) = -\frac{b}{a} \]