Let `tan^(-1)y=tan^(-1)x+tan^(-1)((2x)/(1-x^2))` , where `|x|<1/(sqrt(3))` . Then a value of y is : (1) `(3x-x^3)/(1-3x^2)` (2) `(3x+x^3)/(1-3x^2)` (3) `(3x-x^3)/(1+3x^2)` (4) `(3x+x^3)/(1+3x^2)`

Options

A. `(3x -x^(3))/(1 -3x^(2))`
B. `(3x + x^(3))/(1 - 3x^(2))`
C. `(3x -x^(3))/(1 + 3x^(2))`
D. `(3x + x^(3))/(1 + 3x^(2))`

Explanation

To solve the equation \( \tan^{-1} y = \tan^{-1} x + \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \), where \( |x| < \frac{1}{\sqrt{3}} \), we can use the addition formula for the tangent function.

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \tan^{-1} y = \tan^{-1} x + \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \]

2. **Use the tangent addition formula:**
The formula states that:
\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \]
if \( ab < 1 \).

Here, let \( a = x \) and \( b = \frac{2x}{1 - x^2} \).

3. **Calculate \( ab \):**
\[ ab = x \cdot \frac{2x}{1 - x^2} = \frac{2x^2}{1 - x^2} \]
Since \( |x| < \frac{1}{\sqrt{3}} \), we can verify that \( ab < 1 \).

4. **Apply the addition formula:**
\[ y = \frac{x + \frac{2x}{1 - x^2}}{1 - x \cdot \frac{2x}{1 - x^2}} \]

5. **Simplify the numerator:**
\[ x + \frac{2x}{1 - x^2} = \frac{x(1 - x^2) + 2x}{1 - x^2} = \frac{x - x^3 + 2x}{1 - x^2} = \frac{3x - x^3}{1 - x^2} \]

6. **Simplify the denominator:**
\[ 1 - x \cdot \frac{2x}{1 - x^2} = 1 - \frac{2x^2}{1 - x^2} = \frac{(1 - x^2) - 2x^2}{1 - x^2} = \frac{1 - 3x^2}{1 - x^2} \]

7. **Combine the results:**
\[ y = \frac{\frac{3x - x^3}{1 - x^2}}{\frac{1 - 3x^2}{1 - x^2}} = \frac{3x - x^3}{1 - 3x^2} \]

Thus, the value of \( y \) is:
\[ y = \frac{3x - x^3}{1 - 3x^2} \]

### Final Answer:
The value of \( y \) is \( \frac{3x - x^3}{1 - 3x^2} \), which corresponds to option (1).