Question:

Step 1: Recognize the perfect square pattern in the given equation. We observe that the left side of the equation has the form $a^2 + b^2 + 2ab$, which is a perfect square trinomial. Let us rewrite: $$(\sin^{-1}x)^2 + (\sin^{-1}y)^2 + 2\sin^{-1}x\sin^{-1}y = (\sin^{-1}x + \sin^{-1}y)^2$$ Step 2: Simplify the equation using the perfect square. Since the left side equals $\pi^2$, we have: $$(\sin^{-1}x + \sin^{-1}y)^2 = \pi^2$$ Taking the square root of both sides: $$\sin^{-1}x + \sin^{-1}y = \pm\pi$$ Step 3: Determine the valid range for the sum. Recall that the range of $\sin^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Therefore, the sum $\sin^{-1}x + \sin^{-1}y$ must lie in the interval: $$\left[-\frac{\pi}{2} - \frac{\pi}{2}, \frac{\pi}{2} + \frac{\pi}{2}\right] = [-\pi, \pi]$$ Both values $\pi$ and $-\pi$ are within this range, so both cases are potentially valid. Step 4: Analyze the case where $\sin^{-1}x + \sin^{-1}y = \pi$. For the sum to equal $\pi$, and given that each term is at most $\frac{\pi}{2}$, both must achieve their maximum value: $$\sin^{-1}x = \frac{\pi}{2} \quad \text{and} \quad \sin^{-1}y = \frac{\pi}{2}$$ This gives us $x = 1$ and $y = 1$. Step 5: Analyze the case where $\sin^{-1}x + \sin^{-1}y = -\pi$. For the sum to equal $-\pi$, both must achieve their minimum value: $$\sin^{-1}x = -\frac{\pi}{2} \quad \text{and} \quad \sin^{-1}y = -\frac{\pi}{2}$$ This gives us $x = -1$ and $y = -1$. Step 6: Calculate $x^2 + y^2$ for both cases. For Case 1: $x = 1, y = 1$ $$x^2 + y^2 = 1^2 + 1^2 = 2$$ For Case 2: $x = -1, y = -1$ $$x^2 + y^2 = (-1)^2 + (-1)^2 = 2$$ In both cases, we obtain the same result. **Final Answer:** $x^2 + y^2 = 2$ The answer is **Option 3**. <div class="key-concept"><strong>Key Concept:</strong> Properties of inverse trigonometric functions and algebraic manipulation</div> <div class="trap-box"><strong>Trap:</strong> Not recognizing the perfect square on the left side.</div>