If `(1+sec^(2)pi x). (1+sec^(2)y)= -x^(2)+2x+3` then

Options

A. `x= -1, y= npi, n in I`
B. `x = 0, y = n pi, n in I`
C. `x = 1, y = 2n pi, n in I`
D. `x = 1, y = n pi, n in I`

Explanation

To solve the equation \( (1 + \sec^2(\pi x))(1 + \sec^2(y)) = -x^2 + 2x + 3 \), we will follow these steps:

### Step 1: Understand the given equation
The equation is given as:
\[ (1 + \sec^2(\pi x))(1 + \sec^2(y)) = -x^2 + 2x + 3 \]
We need to analyze both sides of the equation.

### Step 2: Simplify the right-hand side (RHS)
The right-hand side can be rewritten as:
\[ -x^2 + 2x + 3 = -(x^2 - 2x - 3) \]
Factoring the quadratic expression:
\[ x^2 - 2x - 3 = (x - 3)(x + 1) \]
Thus, we have:
\[ -x^2 + 2x + 3 = -((x - 3)(x + 1)) \]

### Step 3: Find the maximum value of the left-hand side (LHS)
The left-hand side is:
\[ 1 + \sec^2(\pi x) \text{ and } 1 + \sec^2(y) \]
Since \(\sec^2(\theta) \geq 1\) for all \(\theta\), we have:
\[ 1 + \sec^2(\pi x) \geq 2 \quad \text{and} \quad 1 + \sec^2(y) \geq 2 \]
Thus, the minimum value of the LHS is:
\[ (1 + \sec^2(\pi x))(1 + \sec^2(y)) \geq 2 \times 2 = 4 \]

### Step 4: Set LHS equal to RHS
Since the minimum value of the LHS is 4, we set:
\[ 4 = -x^2 + 2x + 3 \]
Rearranging gives:
\[ -x^2 + 2x - 1 = 0 \]
Multiplying through by -1:
\[ x^2 - 2x + 1 = 0 \]
Factoring:
\[ (x - 1)^2 = 0 \]
Thus, \(x = 1\).

### Step 5: Substitute \(x = 1\) back into the LHS
Substituting \(x = 1\) into the LHS:
\[ 1 + \sec^2(\pi \cdot 1) = 1 + \sec^2(\pi) = 1 + 1 = 2 \]
So we have:
\[ (1 + \sec^2(\pi))(1 + \sec^2(y)) = 2(1 + \sec^2(y)) \]
Setting this equal to 4:
\[ 2(1 + \sec^2(y)) = 4 \]
Dividing by 2:
\[ 1 + \sec^2(y) = 2 \]
Thus:
\[ \sec^2(y) = 1 \implies y = n\pi \quad (n \in \mathbb{Z}) \]

### Final Solution
The solutions to the equation are:
\[ x = 1 \quad \text{and} \quad y = n\pi \quad (n \text{ is any integer}) \]