Define: `tan^(-1)((3x-x^(3))/(1-3x^(2)))` in terms of `tan^(-1)x`

Explanation

To define \( \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) \) in terms of \( \tan^{-1}(x) \), we can use the addition formula for the tangent inverse function. The formula states that:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]

### Step-by-Step Solution:

1. **Identify the expression**: We need to express \( \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) \).

2. **Use the addition formula**: We can express \( \frac{3x - x^3}{1 - 3x^2} \) in terms of \( \tan^{-1}(x) \). Notice that:
- \( 3x \) can be thought of as \( \tan^{-1}(x) + \tan^{-1}(x) + \tan^{-1}(x) \).

3. **Express \( 3x \) in terms of \( \tan^{-1}(x) \)**:
- Using the addition formula, we can write:
\[ \tan^{-1}(x) + \tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1 - x^2}\right) \]
- Now, add \( \tan^{-1}(x) \) again:
\[ \tan^{-1}\left(\frac{2x}{1 - x^2}\right) + \tan^{-1}(x) = \tan^{-1}\left(\frac{\frac{2x}{1 - x^2} + x}{1 - \frac{2x \cdot x}{1 - x^2}}\right) \]

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

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

6. **Combine the results**:
- Thus, we have:
\[ \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) = \tan^{-1}(x) + \tan^{-1}(x) + \tan^{-1}(x) = 3\tan^{-1}(x) \]

### Final Result:
\[ \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) = 3\tan^{-1}(x) \]