show that `tan^(-1)x+tan^(-1)((2x)/(1-x^2))=tan^(-1)((3x-x^3)/(1-3x^2)),|x|<1/sqrt3`

Explanation

To show that

\[ \tan^{-1}(x) + \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) \quad \text{for } |x| < \frac{1}{\sqrt{3}}, \]

we can use the properties of the tangent function and the addition formula for inverse tangents.

### Step-by-step Solution:

1. **Let \( x = \tan(\theta) \)**:
- Therefore, \( \tan^{-1}(x) = \theta \).

2. **Substituting into the left-hand side (LHS)**:
- The LHS becomes:
\[ \tan^{-1}(x) + \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \theta + \tan^{-1}\left(\frac{2\tan(\theta)}{1 - \tan^2(\theta)}\right). \]

3. **Using the tangent addition formula**:
- Recall that:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}. \]
- Here, let \( a = \theta \) and \( b = \tan^{-1}\left(\frac{2\tan(\theta)}{1 - \tan^2(\theta)}\right) \).
- Then, \( \tan b = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \).

4. **Calculating \( \tan(\theta + b) \)**:
- We have:
\[ \tan(\theta + b) = \frac{\tan(\theta) + \tan(b)}{1 - \tan(\theta) \tan(b)} = \frac{\tan(\theta) + \frac{2\tan(\theta)}{1 - \tan^2(\theta)}}{1 - \tan(\theta) \cdot \frac{2\tan(\theta)}{1 - \tan^2(\theta)}}. \]
- Substituting \( \tan(\theta) = x \):
\[ = \frac{x + \frac{2x}{1 - x^2}}{1 - x \cdot \frac{2x}{1 - x^2}}. \]

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

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

7. **Putting it all together**:
- Thus, we have:
\[ \tan(\theta + b) = \frac{3x - x^3}{1 - 3x^2}. \]

8. **Identifying the right-hand side (RHS)**:
- The RHS is:
\[ \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right). \]

9. **Conclusion**:
- Therefore, we conclude that:
\[ \tan^{-1}(x) + \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right). \]