If `tan^(2) {pi(x+y)}+cot^(2) {pi (x+y)}=1+sqrt((2x)/(1+x^(2)))` where `x, y in R`, then find the least possible value of y.

Explanation

To solve the equation \( \tan^2(\pi(x+y)) + \cot^2(\pi(x+y)) = 1 + \sqrt{\frac{2x}{1+x^2}} \) where \( x, y \in \mathbb{R} \), we will follow these steps:

### Step 1: Analyze the left-hand side (LHS)
Using the identity \( \tan^2 a + \cot^2 a \geq 2 \) (with equality when \( a = \frac{\pi}{4} \)), we can say:
\[ \tan^2(\pi(x+y)) + \cot^2(\pi(x+y)) \geq 2 \]
This gives us our first inequality:
\[ \text{LHS} \geq 2 \]

### Step 2: Analyze the right-hand side (RHS)
Next, we need to analyze the right-hand side:
\[ 1 + \sqrt{\frac{2x}{1+x^2}} \]
To find the maximum value of \( \sqrt{\frac{2x}{1+x^2}} \), we can set \( f(x) = \frac{2x}{1+x^2} \) and find its maximum.

### Step 3: Find the maximum of \( \sqrt{\frac{2x}{1+x^2}} \)
To maximize \( f(x) = \frac{2x}{1+x^2} \), we can take its derivative and set it to zero:
\[ f'(x) = \frac{(1+x^2)(2) - 2x(2x)}{(1+x^2)^2} = \frac{2(1+x^2 - 2x^2)}{(1+x^2)^2} = \frac{2(1-x^2)}{(1+x^2)^2} \]
Setting \( f'(x) = 0 \) gives:
\[ 1 - x^2 = 0 \implies x = 1 \quad \text{(since } x \geq 0 \text{)} \]
Now, substituting \( x = 1 \) into \( f(x) \):
\[ f(1) = \frac{2(1)}{1+1^2} = \frac{2}{2} = 1 \]
Thus, the maximum value of \( \sqrt{\frac{2x}{1+x^2}} \) is \( 1 \).

### Step 4: Substitute back into the RHS
Now substituting back, we find:
\[ 1 + \sqrt{\frac{2x}{1+x^2}} \leq 1 + 1 = 2 \]
This gives us our second inequality:
\[ \text{RHS} \leq 2 \]

### Step 5: Equate LHS and RHS
From the results of Step 1 and Step 4, we have:
\[ \tan^2(\pi(x+y)) + \cot^2(\pi(x+y)) = 2 \]
This means:
\[ \tan^2(\pi(x+y)) + \cot^2(\pi(x+y)) = 2 \quad \text{when } \tan(\pi(x+y)) = 1 \]
This occurs when:
\[ \pi(x+y) = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]
Thus:
\[ x+y = \frac{1}{4} + n \]

### Step 6: Solve for \( y \)
Rearranging gives:
\[ y = \frac{1}{4} + n - x \]
To find the least possible value of \( y \), we need to minimize \( y \). The minimum occurs when \( n \) is minimized (let \( n = -1 \)):
\[ y = \frac{1}{4} - 1 - x = -\frac{3}{4} - x \]
Since \( x \) can take any real value, to minimize \( y \), we can let \( x \) approach \( -\infty \). However, we need to find the least value of \( y \) when \( x = 1 \) (as derived earlier):
\[ y = \frac{1}{4} - 1 = -\frac{3}{4} \]

### Conclusion
Thus, the least possible value of \( y \) is:
\[ \boxed{-\frac{3}{4}} \]