Simplity `tan{1/2sin^(-1)((2x)/(1+x^(2)))+1/2cos^(-1)(1-y^(2))/(1+y^(2))}`, if `xgtygt1`.

Explanation

To simplify the expression \( \tan\left(\frac{1}{2}\sin^{-1}\left(\frac{2x}{1+x^2}\right) + \frac{1}{2}\cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right) \), given that \( x > y > 1 \), we can follow these steps:

### Step 1: Use the identities for inverse trigonometric functions
We start by recognizing that:
- \( x = \tan(\theta) \) implies \( \theta = \tan^{-1}(x) \)
- \( y = \tan(\phi) \) implies \( \phi = \tan^{-1}(y) \)

### Step 2: Substitute \( x \) and \( y \)
Substituting \( x = \tan(\theta) \) and \( y = \tan(\phi) \) into the expression gives:
\[ \tan\left(\frac{1}{2}\sin^{-1}\left(\frac{2\tan(\theta)}{1+\tan^2(\theta)}\right) + \frac{1}{2}\cos^{-1}\left(\frac{1-\tan^2(\phi)}{1+\tan^2(\phi)}\right)\right) \]

### Step 3: Simplify the inverse sine and cosine
Using the double angle identities:
- \( \sin(2\theta) = \frac{2\tan(\theta)}{1+\tan^2(\theta)} \)
- \( \cos(2\phi) = \frac{1-\tan^2(\phi)}{1+\tan^2(\phi)} \)

We can rewrite the expression as:
\[ \tan\left(\frac{1}{2}\sin^{-1}(\sin(2\theta)) + \frac{1}{2}\cos^{-1}(\cos(2\phi))\right) \]

### Step 4: Apply the half-angle identities
Now we can simplify further:
\[ \tan\left(\theta + \phi\right) = \frac{\tan(\theta) + \tan(\phi)}{1 - \tan(\theta)\tan(\phi)} \]
Substituting back \( \tan(\theta) = x \) and \( \tan(\phi) = y \):
\[ \tan(\theta + \phi) = \frac{x + y}{1 - xy} \]

### Step 5: Final expression
Thus, the simplified expression is:
\[ \frac{x + y}{1 - xy} \]

### Summary of Steps
1. Recognize the relationships between \( x, y \) and angles \( \theta, \phi \).
2. Substitute \( x = \tan(\theta) \) and \( y = \tan(\phi) \).
3. Use the identities for sine and cosine to rewrite the expression.
4. Apply the tangent addition formula to combine the angles.
5. Substitute back to express in terms of \( x \) and \( y \).