`tan^(-1)[(C_(1)x-y)/(c_(1)y+x)]+tan^(-1)[(C_(2)-C_(1))/(1+C_(1)C_(2))]..+tan^(-1)[(1)/(c_(n))]` is equal to

Options

A. `tan^(-1)(y)/(x)`
B. `tan^(-1)(x/y)`
C. `-tan^(-1)(x)/(y)`
D. none of these

Explanation

To solve the given problem, we need to evaluate the expression:

\[ \tan^{-1}\left(\frac{C_{1}x - y}{C_{1}y + x}\right) + \tan^{-1}\left(\frac{C_{2} - C_{1}}{1 + C_{1}C_{2}}\right) + \ldots + \tan^{-1\left(\frac{1}{C_{n}}\right)} \]

### Step-by-Step Solution:

1. **Identify the First Term:**
The first term is:
\[ \tan^{-1}\left(\frac{C_{1}x - y}{C_{1}y + x}\right) \]

2. **Rewrite the First Term:**
We can express this in a different form:
\[ \tan^{-1}\left(\frac{C_{1}x - y}{C_{1}y + x}\right) = \tan^{-1}\left(\frac{C_{1}x}{C_{1}y + x} - \frac{y}{C_{1}y + x}\right) \]
This can be simplified further using the identity for the difference of two angles.

3. **Use the Tangent Addition Formula:**
We know that:
\[ \tan^{-1}(A) + \tan^{-1}(B) = \tan^{-1}\left(\frac{A + B}{1 - AB}\right) \]
We can apply this formula to the first term and the subsequent terms.

4. **Evaluate the Subsequent Terms:**
The second term is:
\[ \tan^{-1}\left(\frac{C_{2} - C_{1}}{1 + C_{1}C_{2}}\right) \]
This can be rewritten as:
\[ \tan^{-1}\left(\tan(\theta_2)\right) \text{ where } \theta_2 = \tan^{-1}(C_{2}) - \tan^{-1}(C_{1}) \]

5. **Continue the Pattern:**
Continuing this process for all terms, we notice a pattern where each term cancels with the previous one:
\[ \tan^{-1}(C_{k}) - \tan^{-1}(C_{k-1}) \text{ for } k = 1, 2, \ldots, n \]

6. **Final Simplification:**
After all cancellations, we are left with:
\[ \tan^{-1}\left(\frac{x}{y}\right) \]

7. **Conclusion:**
Therefore, the entire expression simplifies to:
\[ \tan^{-1}\left(\frac{x}{y}\right) \]

### Final Answer:
The final result is:
\[ \tan^{-1}\left(\frac{x}{y}\right) \]