Find the value of
`tan { 1/2 sin^(-1) ((2x)/(1+x^(2))) + 1/2 cos^(-1) ((1-y^(2))/(1+ y^(2)))}`

Answer

To solve the problem, we need to find the value of

\[ \tan \left( \frac{1}{2} \sin^{-1} \left( \frac{2x}{1+x^2} \right) + \frac{1}{2} \cos^{-1} \left( \frac{1-y^2}{1+y^2} \right) \right). \]

### Step 1: Use the known formulas for inverse trigonometric functions

We know the following identities:
- \(\tan \left( \frac{1}{2} \sin^{-1}(z) \right) = \frac{z}{1 + \sqrt{1 - z^2}}\)
- \(\tan \left( \frac{1}{2} \cos^{-1}(z) \right) = \frac{\sqrt{1 - z^2}}{1 + z}\)

### Step 2: Apply the formulas to the given expressions

Let \( z_1 = \frac{2x}{1+x^2} \) and \( z_2 = \frac{1-y^2}{1+y^2} \).

Using the first formula:
\[ \tan \left( \frac{1}{2} \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right) = \frac{\frac{2x}{1+x^2}}{1 + \sqrt{1 - \left( \frac{2x}{1+x^2} \right)^2}}. \]

### Step 3: Simplify the expression for \( \tan \left( \frac{1}{2} \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right) \)

Calculate \( \sqrt{1 - \left( \frac{2x}{1+x^2} \right)^2} \):
\[ 1 - \left( \frac{2x}{1+x^2} \right)^2 = 1 - \frac{4x^2}{(1+x^2)^2} = \frac{(1+x^2)^2 - 4x^2}{(1+x^2)^2} = \frac{1 - 2x^2 + x^4}{(1+x^2)^2} = \frac{(1-x^2)^2}{(1+x^2)^2}. \]
Thus,
\[ \sqrt{1 - \left( \frac{2x}{1+x^2} \right)^2} = \frac{1-x^2}{1+x^2}. \]

Substituting back, we have:
\[ \tan \left( \frac{1}{2} \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right) = \frac{\frac{2x}{1+x^2}}{1 + \frac{1-x^2}{1+x^2}} = \frac{\frac{2x}{1+x^2}}{\frac{2}{1+x^2}} = x. \]

### Step 4: Apply the second formula for \( \tan \left( \frac{1}{2} \cos^{-1} \left( \frac{1-y^2}{1+y^2} \right) \right) \)

Now for \( z_2 = \frac{1-y^2}{1+y^2} \):
\[ \tan \left( \frac{1}{2} \cos^{-1} \left( \frac{1-y^2}{1+y^2} \right) \right) = \frac{\sqrt{1 - \left( \frac{1-y^2}{1+y^2} \right)^2}}{1 + \frac{1-y^2}{1+y^2}}. \]

Calculating \( \sqrt{1 - \left( \frac{1-y^2}{1+y^2} \right)^2} \):
\[ 1 - \left( \frac{1-y^2}{1+y^2} \right)^2 = 1 - \frac{(1-y^2)^2}{(1+y^2)^2} = \frac{(1+y^2)^2 - (1-y^2)^2}{(1+y^2)^2} = \frac{4y^2}{(1+y^2)^2}. \]
Thus,
\[ \sqrt{1 - \left( \frac{1-y^2}{1+y^2} \right)^2} = \frac{2y}{1+y^2}. \]

Substituting back, we have:
\[ \tan \left( \frac{1}{2} \cos^{-1} \left( \frac{1-y^2}{1+y^2} \right) \right) = \frac{\frac{2y}{1+y^2}}{\frac{2}{1+y^2}} = y. \]

### Step 5: Combine the results

Now we can combine the results:
\[ \tan \left( \frac{1}{2} \sin^{-1} \left( \frac{2x}{1+x^2} \right) + \frac{1}{2} \cos^{-1} \left( \frac{1-y^2}{1+y^2} \right) \right) = \frac{x + y}{1 - xy}. \]

Thus, the final answer is:

\[ \tan \left( \frac{1}{2} \sin^{-1} \left( \frac{2x}{1+x^2} \right) + \frac{1}{2} \cos^{-1} \left( \frac{1-y^2}{1+y^2} \right) \right) = \frac{x + y}{1 - xy}. \]