<p><strong>158.</strong> 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 that the left side is a perfect square trinomial: (sin⁻¹x + sin⁻¹y)² = π². Since sin⁻¹x and sin⁻¹y must satisfy the range constraint [-π/2, π/2], their sum equals π, forcing both to equal π/2, which means x = y = 1.
<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> Set up the equation from the perfect square.</p><p>(sin⁻¹x + sin⁻¹y)² = π²</p><p>∴ sin⁻¹x + sin⁻¹y = ±π</p><p><strong>Step 3:</strong> Apply domain constraints.</p><p>Since x, y ∈ [-1, 1], we have sin⁻¹x, sin⁻¹y ∈ [-π/2, π/2]</p><p>Maximum value of sin⁻¹x + sin⁻¹y = π/2 + π/2 = π</p><p>Minimum value = -π/2 - π/2 = -π</p><p>Therefore: sin⁻¹x + sin⁻¹y = π (taking positive value)</p><p><strong>Step 4:</strong> Determine x and y.</p><p>For the sum to equal π with both terms in [-π/2, π/2], both must equal π/2</p><p>sin⁻¹x = π/2 ⟹ x = 1</p><p>sin⁻¹y = π/2 ⟹ y = 1</p><p><strong>Step 5:</strong> Calculate x² + y².</p><p>x² + y² = 1² + 1² = 2</p><p>∴ Answer: A</p>
Correct Answer: A