The formula `"cos"^(-1)(1-x^(2))/(1+x^(2))=2tan^(-1)x` holds only for :

Options

A. `x in R`
B. `|x| le 1`
C. `x in (-1, 1)`
D. `x in [0, +oo]`

Answer

To determine the conditions under which the formula

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

holds, we will follow these steps:

### Step 1: Set up the equation
We start with the left-hand side (LHS) and right-hand side (RHS) of the equation:

\[ \text{LHS} = \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \]
\[ \text{RHS} = 2 \tan^{-1}(x) \]

### Step 2: Substitute \( x = \tan(\theta) \)
To simplify the LHS, we substitute \( x \) with \( \tan(\theta) \):

\[ \text{LHS} = \cos^{-1}\left(\frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)}\right) \]

### Step 3: Use the trigonometric identity
Recall the identity:

\[ \cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} \]

Thus, we can rewrite the LHS:

\[ \text{LHS} = \cos^{-1}(\cos(2\theta)) = 2\theta \]

### Step 4: Relate \( \theta \) to \( x \)
Since \( \theta = \tan^{-1}(x) \), we have:

\[ \text{LHS} = 2 \tan^{-1}(x) \]

This shows that the LHS equals the RHS, confirming that the formula holds.

### Step 5: Determine the conditions
The equality \( \cos^{-1}(\cos(2\theta)) = 2\theta \) holds under certain conditions. Specifically, \( 2\theta \) must be in the range \( [0, \pi] \):

\[ 0 \leq 2\theta \leq \pi \]

This implies:

\[ 0 \leq \theta \leq \frac{\pi}{2} \]

### Step 6: Convert back to \( x \)
Since \( \theta = \tan^{-1}(x) \), we find:

\[ 0 \leq \tan^{-1}(x) \leq \frac{\pi}{2} \]

Taking the tangent of all parts gives:

\[ 0 \leq x < \infty \]

However, since we also need to consider the original expression \( \frac{1 - x^2}{1 + x^2} \) to ensure it remains valid, we also recognize that \( x \) must be within the bounds of the cosine function, specifically:

\[ |x| \leq 1 \]

### Conclusion
Thus, the formula holds for:

\[ x \in [0, 1] \]