If` x in (1,oo) then tan^(-1)((2x)/(1-x^(2)))` equals

Options

A. `2 tan^(-1)x`
B. `-pi+2tan^(-1)x`
C. `pi+2 tan^(-1)x`
D. none of these

Answer

To solve the problem, we need to find the value of \( \tan^{-1}\left(\frac{2x}{1 - x^2}\right) \) for \( x \in (1, \infty) \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression \( \tan^{-1}\left(\frac{2x}{1 - x^2}\right) \). We recognize that this expression resembles the tangent double angle formula:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]
This suggests that we can set \( \tan(\theta) = x \).

2. **Setting Up the Equation**:
If we let \( \theta = \tan^{-1}(x) \), then:
\[ \tan(\theta) = x \]
Therefore, we can express \( \tan(2\theta) \) as:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} = \frac{2x}{1 - x^2} \]

3. **Relating the Angles**:
From the previous step, we can conclude:
\[ \tan^{-1}\left(\frac{2x}{1 - x^2}\right) = 2\theta \]
Since \( \theta = \tan^{-1}(x) \), we have:
\[ \tan^{-1}\left(\frac{2x}{1 - x^2}\right) = 2\tan^{-1}(x) \]

4. **Considering the Range**:
However, we need to consider the range of the function. Since \( x > 1 \), \( \tan^{-1}(x) \) is positive, and thus \( 2\tan^{-1}(x) \) is also positive. But we also need to check the behavior of \( \frac{2x}{1 - x^2} \) when \( x > 1 \):
- The numerator \( 2x \) is positive.
- The denominator \( 1 - x^2 \) is negative (since \( x^2 > 1 \)).
Therefore, \( \frac{2x}{1 - x^2} < 0 \).

5. **Final Adjustment**:
Since \( \tan^{-1}\left(\frac{2x}{1 - x^2}\right) \) must yield a negative value, we can express it as:
\[ \tan^{-1}\left(\frac{2x}{1 - x^2}\right) = -\left(\pi - 2\tan^{-1}(x)\right) = -\pi + 2\tan^{-1}(x) \]

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