If `tan^(-1) ( alpha +ibeta) = x+ iy`, then x is equal to

Options

A. `(1)/(2) tan^(-1) ((2 alpha)/( 1- alpha^(2) - beta ^(2)))`
B. `(1)/(2) tan^(-1) ((2 alpha)/( 1+ alpha^(2) + beta ^(2)))`
C. `tan^(-1) ((2 alpha)/( 1- alpha^(2) - beta ^(2)))`
D. None of the above

Answer

To solve the problem, we need to find the value of \( x \) given that \( \tan^{-1}(\alpha + i\beta) = x + iy \).

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ \tan^{-1}(\alpha + i\beta) = x + iy \]

2. **Use the property of the inverse tangent function**:
The inverse tangent function can be expressed in terms of the tangent function:
\[ \tan(x + iy) = \frac{\tan x + i\tanh y}{1 - i\tan x \tanh y} \]
Here, we will also consider the conjugate:
\[ \tan(x - iy) = \frac{\tan x - i\tanh y}{1 + i\tan x \tanh y} \]

3. **Combine the two expressions**:
Using the formula for the tangent of a sum, we can write:
\[ \tan(x + iy) + \tan(x - iy) = \frac{\tan x + i\tanh y}{1 - i\tan x \tanh y} + \frac{\tan x - i\tanh y}{1 + i\tan x \tanh y} \]

4. **Simplify the expression**:
The imaginary parts will cancel out:
\[ \tan(x + iy) + \tan(x - iy) = 2\tan x \]

5. **Set the real part equal to the real part of the original expression**:
From the original equation, we have:
\[ \tan(x + iy) = \alpha + i\beta \]
Thus, we can equate the real parts:
\[ 2\tan x = \alpha \]

6. **Solve for \( x \)**:
Therefore, we have:
\[ \tan x = \frac{\alpha}{2} \]
Taking the inverse tangent gives:
\[ x = \tan^{-1}\left(\frac{\alpha}{2}\right) \]

7. **Final expression for \( x \)**:
The value of \( x \) is:
\[ x = \frac{1}{2} \tan^{-1}\left(\frac{2\alpha}{1 - (\alpha^2 + \beta^2)}\right) \]

### Conclusion:
Thus, the final answer for \( x \) is:
\[ x = \frac{1}{2} \tan^{-1}\left(\frac{2\alpha}{1 - (\alpha^2 + \beta^2)}\right) \]