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

Options

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

Answer

To solve the problem, we need to evaluate the expression \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) for \( x \in (1, \infty) \).

### Step-by-Step Solution:

1. **Identify the Range of x**:
We are given that \( x \in (1, \infty) \). This means \( x \) is greater than 1.

2. **Use the Known Formula**:
There is a known relationship for the inverse sine function:
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \begin{cases} 2 \tan^{-1}(x) & \text{if } x \in (-1, 1) \\ \pi - 2 \tan^{-1}(x) & \text{if } x > 1 \end{cases} \]
Since \( x > 1 \) in our case, we will use the second part of the formula:
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \pi - 2 \tan^{-1}(x) \]

3. **Write the Final Expression**:
Thus, for \( x \in (1, \infty) \), we have:
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \pi - 2 \tan^{-1}(x) \]

### Conclusion:
The final answer is:
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \pi - 2 \tan^{-1}(x) \]