If `x= cos alpha+ i sin alpha, y = cos beta+ i sin beta,` then prove that `(x-y)/(x+y)``=i (tan )(alpha-beta)/2`

Answer

To prove that \(\frac{x-y}{x+y} = i \tan \left(\frac{\alpha - \beta}{2}\right)\), where \(x = \cos \alpha + i \sin \alpha\) and \(y = \cos \beta + i \sin \beta\), we can follow these steps:

### Step 1: Express \(x\) and \(y\) in exponential form
Using Euler's formula, we can express \(x\) and \(y\) as:
\[ x = e^{i \alpha}, \quad y = e^{i \beta} \]

### Step 2: Substitute \(x\) and \(y\) into the expression
We need to find \(\frac{x-y}{x+y}\):
\[ \frac{x-y}{x+y} = \frac{e^{i \alpha} - e^{i \beta}}{e^{i \alpha} + e^{i \beta}} \]

### Step 3: Factor out \(e^{i \beta}\)
In both the numerator and the denominator, we can factor out \(e^{i \beta}\):
\[ = \frac{e^{i \beta} \left(e^{i (\alpha - \beta)} - 1\right)}{e^{i \beta} \left(e^{i (\alpha - \beta)} + 1\right)} \]
This simplifies to:
\[ = \frac{e^{i (\alpha - \beta)} - 1}{e^{i (\alpha - \beta)} + 1} \]

### Step 4: Use the identity for \(e^{i\theta}\)
Using the identity \(e^{i\theta} = \cos \theta + i \sin \theta\), we can write:
\[ = \frac{\left(\cos(\alpha - \beta) + i \sin(\alpha - \beta)\right) - 1}{\left(\cos(\alpha - \beta) + i \sin(\alpha - \beta)\right) + 1} \]

### Step 5: Simplify the expression
This gives us:
\[ = \frac{\cos(\alpha - \beta) - 1 + i \sin(\alpha - \beta)}{\cos(\alpha - \beta) + 1 + i \sin(\alpha - \beta)} \]

### Step 6: Apply trigonometric identities
Using the identities \(1 - \cos(2\theta) = 2 \sin^2(\theta)\) and \(1 + \cos(2\theta) = 2 \cos^2(\theta)\):
- The numerator becomes:
\[ \cos(\alpha - \beta) - 1 = -2 \sin^2\left(\frac{\alpha - \beta}{2}\right) \]
- The denominator becomes:
\[ \cos(\alpha - \beta) + 1 = 2 \cos^2\left(\frac{\alpha - \beta}{2}\right) \]

### Step 7: Substitute back into the expression
Now substituting these into our expression:
\[ = \frac{-2 \sin^2\left(\frac{\alpha - \beta}{2}\right) + i \sin(\alpha - \beta)}{2 \cos^2\left(\frac{\alpha - \beta}{2}\right) + i \sin(\alpha - \beta)} \]

### Step 8: Factor out common terms
Factoring out \(\sin\left(\frac{\alpha - \beta}{2}\right)\) from the numerator and denominator:
\[ = \frac{-\sin\left(\frac{\alpha - \beta}{2}\right) + i \cos\left(\frac{\alpha - \beta}{2}\right)}{\cos\left(\frac{\alpha - \beta}{2}\right) + i \sin\left(\frac{\alpha - \beta}{2}\right)} \]

### Step 9: Recognize the tangent form
This can be recognized as:
\[ = i \tan\left(\frac{\alpha - \beta}{2}\right) \]

### Conclusion
Thus, we have proven that:
\[ \frac{x-y}{x+y} = i \tan\left(\frac{\alpha - \beta}{2}\right) \]