The number of ordered pair (x, y) satisfying the equation `sin^(2) (x+y)+cos^(2) (x-y)=1` which lie on the circle `x^(2)+y^(2)=pi^(2)` is _________.

Explanation

To solve the problem, we need to find the number of ordered pairs \((x, y)\) that satisfy the equation

\[ \sin^2(x+y) + \cos^2(x-y) = 1 \]

and also lie on the circle

\[ x^2 + y^2 = \pi^2. \]

### Step 1: Simplify the Trigonometric Equation

We start with the equation:

\[ \sin^2(x+y) + \cos^2(x-y) = 1. \]

Using the Pythagorean identity, we know that \(\sin^2 \theta + \cos^2 \theta = 1\). Therefore, we can rewrite the equation as:

\[ \sin^2(x+y) + (1 - \sin^2(x-y)) = 1. \]

This simplifies to:

\[ \sin^2(x+y) = \sin^2(x-y). \]