If `|x| le 1`, then ` 2tan^(-1)x + sin^(-1)((2x)/(1+x^(2)))` is equal to

Options

A. `4 tan^(-1)x`
B. `0`
C. `pi/2`
D. `pi`

Explanation

To solve the problem, we need to find the value of the expression \( 2\tan^{-1}(x) + \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) given that \( |x| \leq 1 \).

### Step-by-step Solution:

1. **Substitute \( x \) with \( \tan(\theta) \)**:
Let \( x = \tan(\theta) \). Since \( |x| \leq 1 \), this implies \( |\theta| \leq \frac{\pi}{4} \).

2. **Calculate \( 2\tan^{-1}(x) \)**:
Using the substitution, we have:
\[ 2\tan^{-1}(x) = 2\tan^{-1}(\tan(\theta)) = 2\theta \]

3. **Calculate \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \)**:
We know from the double angle identity for sine that:
\[ \sin(2\theta) = \frac{2\tan(\theta)}{1+\tan^2(\theta)} = \frac{2x}{1+x^2} \]
Therefore,
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \sin^{-1}(\sin(2\theta)) = 2\theta \]

4. **Combine the results**:
Now, we can combine the two results:
\[ 2\tan^{-1}(x) + \sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\theta + 2\theta = 4\theta \]

5. **Substitute back for \( \theta \)**:
Since \( \theta = \tan^{-1}(x) \), we substitute back:
\[ 4\theta = 4\tan^{-1}(x) \]

6. **Final Result**:
Thus, the expression \( 2\tan^{-1}(x) + \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) simplifies to:
\[ \boxed{4\tan^{-1}(x)} \]