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

Options

A. `2tan^(-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 \( \tan^{-1}\left(\frac{2x}{1-x^2}\right) \) for \( x \in (-\infty, -1) \).

### Step-by-Step Solution:

1. **Identify the Range of x**:
We are given that \( x \) belongs to the interval \( (-\infty, -1) \). This means \( x \) is less than -1.

2. **Use the Inverse Tangent Formula**:
We can use the identity for the tangent of a double angle:
\[ 2\tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \]
This identity holds for \( x \) in the interval \( (-1, 1) \). However, since \( x < -1 \), we need to adjust our approach.

3. **Adjust for the Range of x**:
For \( x < -1 \), we can use the modified version of the formula:
\[ 2\tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) + \pi \]
This means:
\[ \tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2\tan^{-1}(x) - \pi \]

4. **Express the Result**:
Since \( x < -1 \), we can express the result as:
\[ \tan^{-1}\left(\frac{2x}{1-x^2}\right) = -\pi + 2\tan^{-1}(x) \]

5. **Final Answer**:
Therefore, the expression \( \tan^{-1}\left(\frac{2x}{1-x^2}\right) \) for \( x \in (-\infty, -1) \) simplifies to:
\[ \tan^{-1}\left(\frac{2x}{1-x^2}\right) = -\pi + 2\tan^{-1}(x) \]