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 simplify the expression \(\frac{1+C+iS}{1+C-iS}\) given that \(C^2 + S^2 = 1\).

### Step-by-step Solution:

1. **Identify the given condition**: We know that \(C^2 + S^2 = 1\). This implies that \(C\) and \(S\) can be interpreted as \(\cos(\theta)\) and \(\sin(\theta)\), respectively.

2. **Write the expression**: We need to simplify the expression:
\[ \frac{1 + C + iS}{1 + C - iS} \]

3. **Rationalize the denominator**: To simplify, we multiply the numerator and the denominator by the conjugate of the denominator:
\[ \frac{(1 + C + iS)(1 + C + iS)}{(1 + C - iS)(1 + C + iS)} \]

4. **Calculate the denominator**: The denominator simplifies as follows:
\[ (1 + C - iS)(1 + C + iS) = (1 + C)^2 - (iS)^2 = (1 + C)^2 + S^2 \]
Since \(i^2 = -1\), we have:
\[ (1 + C)^2 + S^2 = (1 + C)^2 + (1 - C^2) = 1 + 2C + C^2 + 1 - C^2 = 2 + 2C = 2(1 + C) \]

5. **Calculate the numerator**: The numerator simplifies as follows:
\[ (1 + C + iS)(1 + C + iS) = (1 + C)^2 + 2(1 + C)(iS) + (iS)^2 = (1 + C)^2 - S^2 + 2(1 + C)(iS) \]
Again, substituting \(S^2 = 1 - C^2\):
\[ (1 + C)^2 - (1 - C^2) = (1 + C)^2 - 1 + C^2 = 2C + C^2 + C^2 = 2C + 2C^2 = 2C(1 + C) \]

6. **Combine the results**: Now we can combine the numerator and denominator:
\[ \frac{2C(1 + C)}{2(1 + C)} = C \]

7. **Final result**: Thus, the expression simplifies to:
\[ C + iS \]

### Conclusion:
The final answer is:
\[ C + iS \]