If `C^(2)+S^(2)=1`, then `(1+C+iS)/(1+C-iS)` is equal to

Options

A. `C+iS`
B. `C-iS`
C. `S+iC`
D. `S-iC`

Answer

To solve the problem, we need to find the value of the expression \(\frac{1 + C + iS}{1 + C - iS}\) given that \(C^2 + S^2 = 1\).

### Step-by-step Solution:

1. **Let \( A = \frac{1 + C + iS}{1 + C - iS} \)**

2. **Rationalize the expression**: Multiply the numerator and the denominator by the conjugate of the denominator, which is \(1 + C + iS\).
\[ A = \frac{(1 + C + iS)(1 + C + iS)}{(1 + C - iS)(1 + C + iS)} \]

3. **Calculate the numerator**:
\[ (1 + C + iS)(1 + C + iS) = (1 + C)^2 + 2(1 + C)(iS) + (iS)^2 \]
Expanding this gives:
\[ = (1 + 2C + C^2) + 2(1 + C)(iS) - S^2 \]
Since \(i^2 = -1\), we have:
\[ = 1 + 2C + C^2 - S^2 + 2(1 + C)(iS) \]

4. **Calculate the denominator**:
\[ (1 + C - iS)(1 + C + iS) = (1 + C)^2 - (iS)^2 \]
This simplifies to:
\[ = (1 + 2C + C^2) + S^2 \]

5. **Combine the results**:
The numerator is:
\[ 1 + 2C + C^2 - S^2 + 2(1 + C)(iS) \]
The denominator is:
\[ 1 + 2C + C^2 + S^2 \]

6. **Substituting \(C^2 + S^2 = 1\)**:
Since \(C^2 + S^2 = 1\), we can replace \(C^2 - S^2\) in the numerator:
\[ = 1 + 2C + (C^2 - S^2) + 2(1 + C)(iS) \]
The denominator simplifies to:
\[ = 1 + 2C + 1 = 2 + 2C \]

7. **Final expression**:
Now we have:
\[ A = \frac{(1 + 2C + 1) + 2(1 + C)(iS)}{2 + 2C} \]
Simplifying this gives:
\[ = \frac{2 + 2C + 2(1 + C)(iS)}{2(1 + C)} \]
\[ = \frac{2(1 + C + (1 + C)iS)}{2(1 + C)} \]
Cancelling \(2(1 + C)\) from the numerator and denominator:
\[ A = 1 + iS \]

### Conclusion:
Thus, the value of \(\frac{1 + C + iS}{1 + C - iS}\) is \(C + iS\).