`sec^(2)theta = 4xy/(x+y)^(2)` is true if and only if

Options

A. `x +y ne 0`
B. `x=y`, `x ne 0`
C. `x = y`
D. `x ne 0, y ne 0 `

Answer

To solve the equation \( \sec^2 \theta = \frac{4xy}{(x+y)^2} \) and find the condition under which it is true, we can follow these steps:

### Step 1: Rewrite the equation
Starting with the given equation:
\[ \sec^2 \theta = \frac{4xy}{(x+y)^2} \]

### Step 2: Take the square root
Taking the square root of both sides gives us:
\[ \sec \theta = \frac{2\sqrt{xy}}{x+y} \]

### Step 3: Relate secant to cosine
Since \( \sec \theta = \frac{1}{\cos \theta} \), we can rewrite the equation as:
\[ \cos \theta = \frac{x+y}{2\sqrt{xy}} \]

### Step 4: Analyze the range of cosine
We know that \( \cos \theta \) varies between -1 and 1. Therefore, we have:
\[ -1 \leq \frac{x+y}{2\sqrt{xy}} \leq 1 \]

### Step 5: Apply the Arithmetic Mean-Geometric Mean Inequality (AM-GM)
Using the AM-GM inequality, we know that:
\[ \frac{x+y}{2} \geq \sqrt{xy} \]
This implies:
\[ \frac{x+y}{2\sqrt{xy}} \geq 1 \]
Thus, we have:
\[ \frac{x+y}{2\sqrt{xy}} \geq 1 \]

### Step 6: Combine inequalities
From the previous steps, we have:
1. \( -1 \leq \frac{x+y}{2\sqrt{xy}} \leq 1 \)
2. \( \frac{x+y}{2\sqrt{xy}} \geq 1 \)

The intersection of these inequalities shows that:
\[ \frac{x+y}{2\sqrt{xy}} = 1 \]

### Step 7: Solve for \( x \) and \( y \)
Setting the equation from the intersection:
\[ \frac{x+y}{2\sqrt{xy}} = 1 \]
Multiplying both sides by \( 2\sqrt{xy} \) gives:
\[ x+y = 2\sqrt{xy} \]

### Step 8: Square both sides
Squaring both sides results in:
\[ (x+y)^2 = 4xy \]
Expanding the left side:
\[ x^2 + 2xy + y^2 = 4xy \]
Rearranging gives:
\[ x^2 + y^2 - 2xy = 0 \]

### Step 9: Factor the equation
This can be factored as:
\[ (x-y)^2 = 0 \]
Thus:
\[ x - y = 0 \implies x = y \]

### Conclusion
The condition under which the original equation \( \sec^2 \theta = \frac{4xy}{(x+y)^2} \) is true is:
\[ \boxed{x = y} \]