If `x_(1) = 2 tan^(-1) ((1 + x)/(1 -x)), x_(2) = sin^(-1) ((1 - x^(2))/(1 + x^(2))), " where " x in (0, 1)`, then 2(`x_(1) + x_(2)`) is equal to

Options

A. 0
B. `2pi`
C. `pi`
D. none of these

Explanation

To solve the problem, we need to evaluate \(2(x_1 + x_2)\) where:

- \(x_1 = 2 \tan^{-1} \left( \frac{1 + x}{1 - x} \right)\)
- \(x_2 = \sin^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)\)

where \(x\) is in the interval \((0, 1)\).

### Step 1: Simplifying \(x_1\)

We start with the expression for \(x_1\):

\[ x_1 = 2 \tan^{-1} \left( \frac{1 + x}{1 - x} \right) \]

Using the double angle formula for tangent, we know:

\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]

Let \(\theta = \tan^{-1} \left( \frac{1 + x}{1 - x} \right)\). Then:

\[ \tan(\theta) = \frac{1 + x}{1 - x} \]

Thus,

\[ \tan(2\theta) = \frac{2 \left( \frac{1 + x}{1 - x} \right)}{1 - \left( \frac{1 + x}{1 - x} \right)^2} \]

Calculating the denominator:

\[ 1 - \left( \frac{1 + x}{1 - x} \right)^2 = 1 - \frac{(1 + x)^2}{(1 - x)^2} = \frac{(1 - x)^2 - (1 + x)^2}{(1 - x)^2} \]

Expanding and simplifying:

\[ (1 - x)^2 - (1 + x)^2 = (1 - 2x + x^2) - (1 + 2x + x^2) = -4x \]

Thus,

\[ 1 - \left( \frac{1 + x}{1 - x} \right)^2 = \frac{-4x}{(1 - x)^2} \]

So we have:

\[ \tan(2\theta) = \frac{2 \left( \frac{1 + x}{1 - x} \right)}{\frac{-4x}{(1 - x)^2}} = -\frac{(1 + x)(1 - x)^2}{2x} \]

This expression is complex, so we will use a trigonometric identity instead.

### Step 2: Using the identity for \(x_1\)

We can use the identity:

\[ \tan^{-1} \left( \frac{1 + x}{1 - x} \right) = \tan^{-1} \left( \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) \right) = \frac{\pi}{4} + \frac{x}{2} \]

Thus,

\[ x_1 = 2\left( \frac{\pi}{4} + \frac{x}{2} \right) = \frac{\pi}{2} + x \]

### Step 3: Simplifying \(x_2\)

Now, we simplify \(x_2\):

\[ x_2 = \sin^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \]

Using the identity:

\[ \sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2} \]

We can rewrite \(x_2\) in terms of cosine:

\[ x_2 = \cos^{-1} \left( \frac{x^2 - 1}{x^2 + 1} \right) \]

### Step 4: Adding \(x_1\) and \(x_2\)

Now we add \(x_1\) and \(x_2\):

\[ x_1 + x_2 = \left( \frac{\pi}{2} + x \right) + \sin^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \]

Using the identity:

\[ \sin^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) + \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) = \frac{\pi}{2} \]

Thus, we find:

\[ x_1 + x_2 = \frac{\pi}{2} \]

### Step 5: Final Calculation

Now we calculate \(2(x_1 + x_2)\):

\[ 2(x_1 + x_2) = 2 \cdot \frac{\pi}{2} = \pi \]

### Conclusion

Thus, the final answer is:

\[ \boxed{\pi} \]