If `alpha=2 tan^-1((1+x)/(1-x)) and B= sin^-1((1-x^2)/(1+x^2))` for `0 lt x lt 1` ,then prove that `alpha+beta =pi` , .

Explanation

To prove that \( \alpha + \beta = \pi \), where

\[ \alpha = 2 \tan^{-1}\left(\frac{1+x}{1-x}\right) \]
and
\[ \beta = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right) \]

for \( 0 < x < 1 \), we will follow these steps:

### Step 1: Substitute \( x = \tan \theta \)

Let \( x = \tan \theta \). Then, we can express \( \alpha \) and \( \beta \) in terms of \( \theta \).

### Step 2: Simplify \( \alpha \)

Substituting \( x = \tan \theta \) into \( \alpha \):

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

Using the tangent addition formula, we have:

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

Thus,

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

### Step 3: Simplify \( \beta \)

Now, substituting \( x = \tan \theta \) into \( \beta \):

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

Using the identity for sine:

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

Thus,

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

### Step 4: Add \( \alpha \) and \( \beta \)

Now, we can add \( \alpha \) and \( \beta \):

\[ \alpha + \beta = \left(\frac{\pi}{2} + 2\theta\right) + \left(\frac{\pi}{2} - 2\theta\right) \]

The \( 2\theta \) terms cancel out:

\[ \alpha + \beta = \frac{\pi}{2} + \frac{\pi}{2} = \pi \]

### Conclusion

Thus, we have proved that:

\[ \alpha + \beta = \pi \]