If ` x in R`, then `f(x)=sin^(-1)((2x)/(1+x^(2)))` is equal to

Options

A. `2 tan^(-1)x`
B. `{{:( -pi-2tan^(-1)x, -oo lt x lt -1),(2 tan^(-1)x , -1 le x le 1),(pi -2 tan^(-1)x, 1 lt x lt oo):}`
C. `{{:( -pi-2tan^(-1)x, -oo lt x lt -1),(2 tan^(-1)x , -1 le x le 1),(pi -2 tan^(-1)x, 1 lt x lt oo):}`
D. `{{:( -pi+2tan^(-1)x, -oo lt x le -1),(2 tan^(-1)x , -1 lt x lt 1),(pi -2 tan^(-1)x, 1 le x lt oo):}`

Answer

To solve the problem, we need to express the function \( f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) in terms of \( \tan^{-1}(x) \).

### Step-by-Step Solution:

1. **Recognize the Identity**:
We know from trigonometric identities that:
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}(x) \]
This identity holds true for \( x \) in the range of \( -1 \leq x \leq 1 \).

2. **Determine the Range for \( \tan^{-1}(x) \)**:
- For \( x \in (-1, 1) \):
\[ f(x) = 2\tan^{-1}(x) \]

3. **Consider Values Outside the Range**:
- For \( x > 1 \):
The function can be expressed as:
\[ f(x) = \pi - 2\tan^{-1}(x) \]
- For \( x < -1 \):
The function can be expressed as:
\[ f(x) = -\pi - 2\tan^{-1}(x) \]

4. **Summarize the Function**:
Therefore, we can summarize \( f(x) \) as:
\[ f(x) = \begin{cases} 2\tan^{-1}(x) & \text{if } -1 < x < 1 \\ \pi - 2\tan^{-1}(x) & \text{if } x \geq 1 \\ -\pi - 2\tan^{-1}(x) & \text{if } x \leq -1 \end{cases} \]

### Final Answer:
Thus, the function \( f(x) \) can be expressed as:
\[ f(x) = \begin{cases} 2\tan^{-1}(x) & \text{if } -1 < x < 1 \\ \pi - 2\tan^{-1}(x) & \text{if } x \geq 1 \\ -\pi - 2\tan^{-1}(x) & \text{if } x \leq -1 \end{cases} \]