If `|cos^(-1) ((1 -x^(2))/(1 + x^(2)))| lt (pi)/(3)`, then

Options

A. `x in [-(1)/(3), (1)/(sqrt3)]`
B. `x in (-(1)/(sqrt3), (1)/(sqrt3))`
C. `x in (0, (1)/(sqrt3))`
D. none of these

Explanation

To solve the inequality \( |\cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)| < \frac{\pi}{3} \), we will follow these steps:

### Step 1: Remove the Absolute Value
Since the expression inside the absolute value can take values from \(0\) to \(\pi\), we can rewrite the inequality without the absolute value:
\[ -\frac{\pi}{3} < \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) < \frac{\pi}{3} \]

### Step 2: Adjust the Range of the Inverse Cosine
The range of \(\cos^{-1}(y)\) is from \(0\) to \(\pi\). Therefore, we can replace \(-\frac{\pi}{3}\) with \(0\):
\[ 0 < \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) < \frac{\pi}{3} \]

### Step 3: Apply the Cosine Function
To eliminate the inverse cosine, we take the cosine of all parts of the inequality:
\[ \cos(0) > \frac{1 - x^2}{1 + x^2} > \cos\left(\frac{\pi}{3}\right) \]
This simplifies to:
\[ 1 > \frac{1 - x^2}{1 + x^2} > \frac{1}{2} \]

### Step 4: Solve the Inequalities
Now we will solve the two inequalities separately.

#### Inequality 1: \(1 > \frac{1 - x^2}{1 + x^2}\)
Cross-multiplying gives:
\[ 1(1 + x^2) > 1 - x^2 \]
This simplifies to:
\[ 1 + x^2 > 1 - x^2 \]
\[ 2x^2 > 0 \quad \Rightarrow \quad x^2 > 0 \]
This means \(x \neq 0\).

#### Inequality 2: \(\frac{1 - x^2}{1 + x^2} > \frac{1}{2}\)
Cross-multiplying gives:
\[ 2(1 - x^2) > 1 + x^2 \]
This simplifies to:
\[ 2 - 2x^2 > 1 + x^2 \]
\[ 2 - 1 > 2x^2 + x^2 \]
\[ 1 > 3x^2 \quad \Rightarrow \quad x^2 < \frac{1}{3} \]
This means:
\[ -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}} \]

### Step 5: Combine the Results
From the inequalities, we have:
1. \(x^2 > 0\) implies \(x \neq 0\).
2. \(x^2 < \frac{1}{3}\) implies \(x\) is in the interval \(-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}\).

Thus, the final solution is:
\[ x \in \left(-\frac{1}{\sqrt{3}}, 0\right) \cup \left(0, \frac{1}{\sqrt{3}}\right) \]