If `0 lt x lt 1 then tan^(-1) (2x)/(1-x^(2))` equals

Options

A. `2 tan^(-1)x`
B. `-pi+2tan^(-1)x`
C. `pi+2tan^(-1)x`
D. none of these

Answer

To solve the problem, we need to find the value of \( \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \) for \( 0 < x < 1 \).

### Step-by-Step Solution:

1. **Recall the Identity for Tangent Addition**:
We know that:
\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \]
for \( ab < 1 \).

2. **Set \( a = x \) and \( b = x \)**:
Using the identity, we can express \( \tan^{-1} x + \tan^{-1} x \):
\[ \tan^{-1} x + \tan^{-1} x = \tan^{-1} \left( \frac{x + x}{1 - x \cdot x} \right) = \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \]

3. **Apply the Identity**:
From the above, we have:
\[ \tan^{-1} \left( \frac{2x}{1 - x^2} \right) = \tan^{-1} x + \tan^{-1} x = 2 \tan^{-1} x \]

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

### Final Answer:
The value of \( \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \) is \( 2 \tan^{-1} x \).

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