If ` a = cos alpha + i sin alpha, b = cos beta + i sin beta , " then " ( a- b)/( a+ b) =`

Options

A. `itan"" (alpha - beta)/( 2)`
B. `icos""(alpha - beta)/(2)`
C. `tan"" (alpha- beta)/(2)`
D. `cot"" (alpha - beta)/(2)`

Answer

To solve the problem, we need to simplify the expression \(\frac{a - b}{a + b}\) where \(a = \cos \alpha + i \sin \alpha\) and \(b = \cos \beta + i \sin \beta\).

### Step-by-step Solution:

1. **Identify the expressions for \(a\) and \(b\)**:
\[ a = \cos \alpha + i \sin \alpha, \quad b = \cos \beta + i \sin \beta \]

2. **Calculate \(a - b\)**:
\[ a - b = (\cos \alpha - \cos \beta) + i(\sin \alpha - \sin \beta) \]

3. **Use the trigonometric identities**:
We can use the identities for the difference of cosines and sines:
\[ \cos \alpha - \cos \beta = -2 \sin\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right) \]
\[ \sin \alpha - \sin \beta = 2 \cos\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right) \]

Thus, we can rewrite \(a - b\):
\[ a - b = -2 \sin\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right) + i \cdot 2 \cos\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right) \]
\[ = 2 \sin\left(\frac{\alpha - \beta}{2}\right) \left(-\sin\left(\frac{\alpha + \beta}{2}\right) + i \cos\left(\frac{\alpha + \beta}{2}\right)\right) \]

4. **Calculate \(a + b\)**:
\[ a + b = (\cos \alpha + \cos \beta) + i(\sin \alpha + \sin \beta) \]
Using the sum identities:
\[ \cos \alpha + \cos \beta = 2 \cos\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) \]
\[ \sin \alpha + \sin \beta = 2 \sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) \]

Therefore, we can write:
\[ a + b = 2 \cos\left(\frac{\alpha - \beta}{2}\right) \left(\cos\left(\frac{\alpha + \beta}{2}\right) + i \sin\left(\frac{\alpha + \beta}{2}\right)\right) \]
\[ = 2 \cos\left(\frac{\alpha - \beta}{2}\right) e^{i \frac{\alpha + \beta}{2}} \]

5. **Combine the results**:
Now we can substitute \(a - b\) and \(a + b\) into the original expression:
\[ \frac{a - b}{a + b} = \frac{2 \sin\left(\frac{\alpha - \beta}{2}\right) \left(-\sin\left(\frac{\alpha + \beta}{2}\right) + i \cos\left(\frac{\alpha + \beta}{2}\right)\right)}{2 \cos\left(\frac{\alpha - \beta}{2}\right) e^{i \frac{\alpha + \beta}{2}}} \]
\[ = \frac{\sin\left(\frac{\alpha - \beta}{2}\right)}{\cos\left(\frac{\alpha - \beta}{2}\right)} \cdot \frac{-\sin\left(\frac{\alpha + \beta}{2}\right) + i \cos\left(\frac{\alpha + \beta}{2}\right)}{e^{i \frac{\alpha + \beta}{2}}} \]

6. **Simplify the expression**:
The expression simplifies to:
\[ = \tan\left(\frac{\alpha - \beta}{2}\right) \cdot e^{-i \frac{\alpha + \beta}{2}} \]

### Final Result:
Thus, the final result is:
\[ \frac{a - b}{a + b} = i \tan\left(\frac{\alpha - \beta}{2}\right) \]