1. **State the problem:** Prove or verify the identity $$(\sin A + \cos A)^2 + (\sin A - \cos A)^2 = 2.$$\n\n2. **Recall the formula:** The square of a sum and difference are given by:\n$$ (x+y)^2 = x^2 + 2xy + y^2 $$\nand\n$$ (x-y)^2 = x^2 - 2xy + y^2.$$\n\n3. **Apply the formula:** Let $x = \sin A$ and $y = \cos A$. Then:\n$$(\sin A + \cos A)^2 = \sin^2 A + 2\sin A \cos A + \cos^2 A,$$\n$$(\sin A - \cos A)^2 = \sin^2 A - 2\sin A \cos A + \cos^2 A.$$\n\n4. **Add the two expressions:**\n$$ (\sin A + \cos A)^2 + (\sin A - \cos A)^2 = (\sin^2 A + 2\sin A \cos A + \cos^2 A) + (\sin^2 A - 2\sin A \cos A + \cos^2 A).$$\n\n5. **Simplify:** The $2\sin A \cos A$ and $-2\sin A \cos A$ cancel out, so we have:\n$$ = \sin^2 A + \cos^2 A + \sin^2 A + \cos^2 A = 2(\sin^2 A + \cos^2 A).$$\n\n6. **Use the Pythagorean identity:**\n$$ \sin^2 A + \cos^2 A = 1,$$\nso\n$$ 2(\sin^2 A + \cos^2 A) = 2 \times 1 = 2.$$\n\n**Final answer:**\n$$(\sin A + \cos A)^2 + (\sin A - \cos A)^2 = 2.$$