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

Options

A. `2 tan^(-1)x`
B. `{{:(2 tan^(-1)x , x ge 0),(-2 tan^(-1)x, x le 0):}`
C. `{{:(pi+2tan^(-1)x, x ge 0),(-pi+ 2 tan^(-1) x, x le0):}`
D. none of these

Answer

To solve the problem, we need to evaluate the function \( f(x) = \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \).

### Step-by-Step Solution:

1. **Recognize the Formula**:
We start with the expression \( \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \). There is a known trigonometric identity that states:
\[ \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) = 2 \tan^{-1}(x) \]
This identity holds true for all \( x \in \mathbb{R} \).

2. **Apply the Formula**:
Based on the identity, we can directly substitute:
\[ f(x) = 2 \tan^{-1}(x) \]

3. **Consider the Domain**:
The function \( \tan^{-1}(x) \) is defined for all real numbers. Thus, \( f(x) = 2 \tan^{-1}(x) \) is also defined for all \( x \in \mathbb{R} \).

4. **Piecewise Definition**:
We can express \( f(x) \) in a piecewise manner based on the sign of \( x \):
- For \( x \geq 0 \):
\[ f(x) = 2 \tan^{-1}(x) \]
- For \( x < 0 \):
\[ f(x) = -2 \tan^{-1}(-x) = -2 \tan^{-1}(x) \text{ (since } \tan^{-1}(-x) = -\tan^{-1}(x)\text{)} \]

5. **Final Expression**:
Therefore, we can summarize the function as:
\[ f(x) = \begin{cases} 2 \tan^{-1}(x) & \text{if } x \geq 0 \\ -2 \tan^{-1}(-x) & \text{if } x < 0 \end{cases} \]

### Final Answer:
Thus, the function \( f(x) = \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \) is equal to \( 2 \tan^{-1}(x) \) for \( x \geq 0 \) and \( -2 \tan^{-1}(-x) \) for \( x < 0 \).