If `a = cos theta + i sin theta, ` then `( 1 + a)/( 1 -a )` is equal to

Options

A. `cot (( theta)/(2))`
B. `cot theta `
C. `i cot "" ( theta)/(2)`
D. `i tan "" ( theta)/(2)`

Answer

To solve the problem, we need to simplify the expression \((1 + a)/(1 - a)\) where \(a = \cos \theta + i \sin \theta\).

### Step-by-step Solution:

1. **Substitute \(a\)**:
\[ a = \cos \theta + i \sin \theta \]
Therefore, we can write:
\[ 1 + a = 1 + \cos \theta + i \sin \theta \]
\[ 1 - a = 1 - (\cos \theta + i \sin \theta) = 1 - \cos \theta - i \sin \theta \]

2. **Rewrite the expression**:
\[ \frac{1 + a}{1 - a} = \frac{1 + \cos \theta + i \sin \theta}{1 - \cos \theta - i \sin \theta} \]

3. **Multiply the numerator and denominator by the conjugate of the denominator**:
The conjugate of \(1 - \cos \theta - i \sin \theta\) is \(1 - \cos \theta + i \sin \theta\).
\[ = \frac{(1 + \cos \theta + i \sin \theta)(1 - \cos \theta + i \sin \theta)}{(1 - \cos \theta - i \sin \theta)(1 - \cos \theta + i \sin \theta)} \]

4. **Simplify the denominator**:
Using the difference of squares:
\[ (1 - \cos \theta)^2 + \sin^2 \theta = 1 - 2\cos \theta + \cos^2 \theta + \sin^2 \theta = 1 - 2\cos \theta + 1 = 2(1 - \cos \theta) \]

5. **Expand the numerator**:
\[ (1 + \cos \theta + i \sin \theta)(1 - \cos \theta + i \sin \theta) = (1 + \cos \theta)(1 - \cos \theta) + i \sin \theta(1 + \cos \theta) + i \sin \theta(1 - \cos \theta) - \sin^2 \theta \]
\[ = 1 - \cos^2 \theta + i \sin \theta(1 + \cos \theta - 1 + \cos \theta) - \sin^2 \theta \]
\[ = \sin^2 \theta + 2i \sin \theta \cos \theta \]

6. **Combine results**:
The numerator simplifies to:
\[ \sin^2 \theta + 2i \sin \theta \cos \theta \]
Thus, we have:
\[ \frac{\sin^2 \theta + 2i \sin \theta \cos \theta}{2(1 - \cos \theta)} \]

7. **Final simplification**:
We can express \(\sin^2 \theta\) as \(1 - \cos^2 \theta\):
\[ = \frac{(1 - \cos^2 \theta) + 2i \sin \theta \cos \theta}{2(1 - \cos \theta)} \]

8. **Recognizing trigonometric identities**:
Using \(1 - \cos^2 \theta = \sin^2 \theta\) and simplifying further leads to:
\[ = \frac{i \sin 2\theta}{2(1 - \cos \theta)} \]

### Conclusion:
The final result is:
\[ \frac{i \sin 2\theta}{2(1 - \cos \theta)} \]