`sin^2 theta =(x+y)^2/(4xy)` where x,y`in`R gives `theta` if and only if

Options

A. `a+y=0`
B. `x=y`
C. `|x|=|y|ne0`
D. none of these

Answer

To solve the equation \( \sin^2 \theta = \frac{(x+y)^2}{4xy} \) where \( x, y \in \mathbb{R} \), we need to find the conditions under which this equation holds true.

### Step-by-Step Solution:

1. **Understand the Range of \( \sin^2 \theta \)**:
\[ 0 \leq \sin^2 \theta \leq 1 \]
This means that the right-hand side of the equation must also fall within this range.

2. **Set Up the Inequality**:
Since \( \sin^2 \theta \) must be between 0 and 1, we can write:
\[ 0 \leq \frac{(x+y)^2}{4xy} \leq 1 \]

3. **Analyze the Lower Bound**:
The expression \( \frac{(x+y)^2}{4xy} \) is non-negative if \( xy > 0 \) (both \( x \) and \( y \) must be either positive or negative). Thus, we need:
\[ xy > 0 \]

4. **Analyze the Upper Bound**:
To satisfy the upper bound, we need:
\[ (x+y)^2 \leq 4xy \]
Rearranging gives:
\[ (x+y)^2 - 4xy \leq 0 \]

5. **Expand the Left Side**:
\[ x^2 + 2xy + y^2 - 4xy \leq 0 \]
Simplifying this results in:
\[ x^2 - 2xy + y^2 \leq 0 \]

6. **Recognize the Perfect Square**:
The expression \( x^2 - 2xy + y^2 \) can be rewritten as:
\[ (x - y)^2 \leq 0 \]

7. **Conclude the Condition**:
Since a square is always non-negative, \( (x - y)^2 \leq 0 \) implies:
\[ (x - y)^2 = 0 \implies x - y = 0 \implies x = y \]

8. **Final Condition**:
We also need to ensure that \( x \) and \( y \) are not zero, as this would make the original expression undefined. Therefore, we conclude:
\[ x = y \quad \text{and} \quad x \neq 0 \]

### Summary of the Condition:
The condition for \( \sin^2 \theta = \frac{(x+y)^2}{4xy} \) to hold true is:
\[ x = y \quad \text{and} \quad x \neq 0 \]