<p>If \((\sin^{-1}x)^2 + (\sin^{-1}y)^2 + 2\sin^{-1}x\sin^{-1}y = \pi^2\), then \(x^2 + y^2\) is equal to:</p>
Step-by-Step Solution
Key Concept: Recognize the equation as a perfect square: (sin⁻¹x + sin⁻¹y)² = π², which means sin⁻¹x + sin⁻¹y = ±π. Since sin⁻¹ has range [-π/2, π/2], the only feasible solution is sin⁻¹x + sin⁻¹y = π (requiring both x and y to be positive with specific values).
<p><strong>Step 1:</strong> Recognize the left side as a perfect square trinomial.</p><p>(sin⁻¹x)² + (sin⁻¹y)² + 2sin⁻¹x·sin⁻¹y = (sin⁻¹x + sin⁻¹y)²</p><p><strong>Step 2:</strong> Therefore: (sin⁻¹x + sin⁻¹y)² = π²</p><p>This gives: sin⁻¹x + sin⁻¹y = ±π</p><p><strong>Step 3:</strong> Since the range of sin⁻¹ is [-π/2, π/2], for x, y ∈ [-1,1], we have sin⁻¹x + sin⁻¹y ∈ [-π, π].</p><p><strong>Step 4:</strong> For the equation to hold with equality π, we need sin⁻¹x = π/2 and sin⁻¹y = π/2, giving x = 1 and y = 1.</p><p>Verification: (π/2)² + (π/2)² + 2(π/2)(π/2) = π²/4 + π²/4 + π²/2 = π² ✓</p><p><strong>Step 5:</strong> Calculate x² + y² = 1² + 1² = 2</p><p>∴ Answer: x² + y² = <strong>2</strong></p>
Correct Answer: A