If `sin^(2)theta=(x^(2)+y^(2)+1)/(2x)`. Find the value of x and y.

Answer

To solve the equation \( \sin^2 \theta = \frac{x^2 + y^2 + 1}{2x} \), we need to find the values of \( x \) and \( y \).

### Step 1: Analyze the given equation
Since \( \sin^2 \theta \) is always between 0 and 1 (inclusive), we can set up the following inequalities:
\[ 0 < \frac{x^2 + y^2 + 1}{2x} < 1 \]

### Step 2: Multiply through by \( 2x \)
Since \( x > 0 \) (as \( x \) cannot be zero because it is in the denominator), we can multiply the entire inequality by \( 2x \):
\[ 0 < x^2 + y^2 + 1 < 2x \]

### Step 3: Rearrange the inequalities
From the right side of the inequality, we can rearrange it:
\[ x^2 + y^2 + 1 < 2x \]
This can be rewritten as:
\[ x^2 - 2x + y^2 + 1 < 0 \]

### Step 4: Complete the square for \( x \)
We can complete the square for the \( x \) terms:
\[ (x - 1)^2 + y^2 < 0 \]

### Step 5: Analyze the inequality
The expression \( (x - 1)^2 + y^2 < 0 \) implies that both \( (x - 1)^2 \) and \( y^2 \) must be equal to zero because squares of real numbers are non-negative. Therefore, we have:
\[ (x - 1)^2 = 0 \quad \text{and} \quad y^2 = 0 \]

### Step 6: Solve for \( x \) and \( y \)
From \( (x - 1)^2 = 0 \), we find:
\[ x - 1 = 0 \implies x = 1 \]
From \( y^2 = 0 \), we find:
\[ y = 0 \]

### Conclusion
Thus, the values of \( x \) and \( y \) are:
\[ \boxed{x = 1, y = 0} \]