If `theta` is the amplitude of `(a+ib)/(a-ib)`, then `tantheta` equals:

Options

A. `(a)/(a^(2)+b^(2))`
B. `(b)/(a^(2)+b^(2))`
C. `(2ab)/(a^(2)-b^(2))`
D. `(2a)/(a^(2)+b^(2))`

Answer

To solve the problem, we need to find the value of \( \tan \theta \) where \( \theta \) is the amplitude (or argument) of the complex number \( \frac{a + ib}{a - ib} \).

### Step-by-Step Solution:

1. **Identify the Complex Numbers**:
Let \( z_1 = a + ib \) and \( z_2 = a - ib \).

2. **Use the Property of Argument**:
The argument of a quotient of two complex numbers can be expressed as:
\[ \text{arg}\left(\frac{z_1}{z_2}\right) = \text{arg}(z_1) - \text{arg}(z_2) \]
Therefore, we have:
\[ \theta = \text{arg}(z_1) - \text{arg}(z_2) \]

3. **Find the Arguments**:
- For \( z_1 = a + ib \):
\[ \text{arg}(z_1) = \tan^{-1}\left(\frac{b}{a}\right) \]
- For \( z_2 = a - ib \):
\[ \text{arg}(z_2) = \tan^{-1}\left(\frac{-b}{a}\right) = -\tan^{-1}\left(\frac{b}{a}\right) \]

4. **Substituting the Arguments**:
Now substituting these into the equation for \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{b}{a}\right) - \left(-\tan^{-1}\left(\frac{b}{a}\right)\right) = \tan^{-1}\left(\frac{b}{a}\right) + \tan^{-1}\left(\frac{b}{a}\right) = 2\tan^{-1}\left(\frac{b}{a}\right) \]

5. **Using the Double Angle Formula for Tangent**:
We know that:
\[ \tan(2\alpha) = \frac{2\tan(\alpha)}{1 - \tan^2(\alpha)} \]
Here, let \( \alpha = \tan^{-1}\left(\frac{b}{a}\right) \), so:
\[ \tan(\alpha) = \frac{b}{a} \]

6. **Substituting into the Double Angle Formula**:
Now substituting \( \tan(\alpha) \) into the double angle formula:
\[ \tan(\theta) = \tan(2\alpha) = \frac{2\tan(\alpha)}{1 - \tan^2(\alpha)} = \frac{2\left(\frac{b}{a}\right)}{1 - \left(\frac{b}{a}\right)^2} \]

7. **Simplifying the Expression**:
Simplifying the right-hand side:
\[ \tan(\theta) = \frac{2\frac{b}{a}}{\frac{a^2 - b^2}{a^2}} = \frac{2b}{a} \cdot \frac{a^2}{a^2 - b^2} = \frac{2ab}{a^2 - b^2} \]

### Final Result:
Thus, we find that:
\[ \tan \theta = \frac{2ab}{a^2 - b^2} \]