` 2 tan^(-1) x = sin^(-1) ((2x)/(1+x^(2))) , 1 le x le 1 `

Answer

To solve the equation \( 2 \tan^{-1} x = \sin^{-1} \left( \frac{2x}{1+x^2} \right) \) for \( -1 \leq x \leq 1 \), we can follow these steps:

### Step 1: Understand the relationship between \( \tan^{-1} \) and \( \sin^{-1} \)

We know from trigonometric identities that:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]
If we let \( \theta = \tan^{-1}(x) \), then \( \tan(\theta) = x \). Thus, we can express \( 2 \tan^{-1}(x) \) as:
\[ \tan(2\tan^{-1}(x)) = \frac{2x}{1 - x^2} \]

### Step 2: Set up the equation

From the above relationship, we can rewrite our original equation:
\[ \tan(2\tan^{-1}(x)) = \frac{2x}{1 - x^2} \]
This means:
\[ \sin^{-1}\left( \frac{2x}{1+x^2} \right) = \tan(2\tan^{-1}(x)) \]

### Step 3: Use the identity for \( \sin^{-1} \)

We know that:
\[ \sin(\sin^{-1}(y)) = y \]
Thus, we can equate:
\[ \sin(2\tan^{-1}(x)) = \frac{2x}{1+x^2} \]

### Step 4: Use the double angle formula for sine

The double angle formula states:
\[ \sin(2\theta) = 2\sin(\theta)\cos(\theta) \]
Using \( \theta = \tan^{-1}(x) \):
- \( \sin(\theta) = \frac{x}{\sqrt{1+x^2}} \)
- \( \cos(\theta) = \frac{1}{\sqrt{1+x^2}} \)

Thus:
\[ \sin(2\tan^{-1}(x)) = 2 \cdot \frac{x}{\sqrt{1+x^2}} \cdot \frac{1}{\sqrt{1+x^2}} = \frac{2x}{1+x^2} \]

### Step 5: Verify the equality

We see that:
\[ \sin(2\tan^{-1}(x)) = \frac{2x}{1+x^2} \]
This confirms that our original equation holds true.

### Step 6: Conclusion

Since the equation holds true for all \( x \) in the interval \( -1 \leq x \leq 1 \), we conclude that the solution is valid for this entire range.