If `-oo lt x le 0 then cos ^(-1)((1-x^(2))/(1+x^(2)))`equals

Options

A. `2tan^(-1)x`
B. `-2 tan^(-1)x`
C. `pi-2 tan^(-1)x`
D. `pi+2tan^(-1)x`

Answer

To solve the problem, we need to find the value of \( \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \) for \( x \) in the range \( (-\infty, 0] \).

### Step-by-Step Solution:

1. **Identify the Range of x**:
We know that \( x \) is in the interval \( (-\infty, 0] \).

2. **Use the Inverse Trigonometric Identity**:
We can use the identity:
\[ \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) = \begin{cases} 2 \tan^{-1}(x) & \text{if } x \geq 0 \\ -2 \tan^{-1}(x) & \text{if } x < 0 \end{cases} \]

3. **Apply the Identity for \( x < 0 \)**:
Since \( x \) is less than 0 in our case, we will use the second part of the identity:
\[ \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) = -2 \tan^{-1}(x) \]

4. **Final Result**:
Therefore, for \( x \in (-\infty, 0] \):
\[ \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) = -2 \tan^{-1}(x) \]

### Conclusion:
The value of \( \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \) when \( x \) is in the range \( (-\infty, 0] \) is \( -2 \tan^{-1}(x) \).