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

Options

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

Answer

To solve the problem \( \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \) for \( 0 \leq x < \infty \), we can use the known relationship between inverse trigonometric functions.

### Step-by-Step Solution:

1. **Recognize the relationship**: We know that for \( x \geq 0 \):
\[ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2\tan^{-1}(x) \]
This is a standard result that can be derived from the definitions of the inverse trigonometric functions.

2. **Apply the relationship**: Since \( x \) is in the range \( [0, \infty) \), we can directly apply the relationship:
\[ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2\tan^{-1}(x) \]

3. **Conclusion**: Therefore, the final answer is:
\[ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2\tan^{-1}(x) \]

### Final Answer:
\[ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2\tan^{-1}(x) \]