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 solve the equation \(\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2\tan^{-1}(x)\) and determine the values or range of \(x\) for which this holds, we can follow these steps:

### Step 1: Substitute \(x\) with \(\tan(\theta)\)
Let \(x = \tan(\theta)\). Then we can rewrite the equation as:
\[ \cos^{-1}\left(\frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)}\right) = 2\tan^{-1}(\tan(\theta)) \]

### Step 2: Use the identity for \(\cos(2\theta)\)
Recall the trigonometric identity:
\[ \cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} \]
Thus, we can rewrite the left-hand side:
\[ \cos^{-1}(\cos(2\theta)) = 2\theta \]

### Step 3: Determine the range for \(2\theta\)
The equation \(\cos^{-1}(\cos(2\theta)) = 2\theta\) holds true when:
\[ 2\theta \in [0, \pi] \]
This means:
\[ 0 \leq 2\theta \leq \pi \]
Dividing by 2 gives:
\[ 0 \leq \theta \leq \frac{\pi}{2} \]

### Step 4: Relate \(\theta\) back to \(x\)
Since \(\theta = \tan^{-1}(x)\), we can express the range of \(\theta\) in terms of \(x\):
\[ 0 \leq \tan^{-1}(x) \leq \frac{\pi}{2} \]

### Step 5: Solve for \(x\)
The range of \(\tan^{-1}(x)\) implies:
\[ x \geq 0 \]
Thus, \(x\) must be in the range:
\[ x \in [0, \infty) \]

### Final Answer
The formula \(\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2\tan^{-1}(x)\) holds only for:
\[ x \in [0, \infty) \]