<p>Find the sum of squares of all values of <em>x</em> satisfying the equation \(2\tan^{-1}x = \dfrac{\pi}{2} - 2(\pi - 2\tan^{-1}x)\) (considering all cases based on the range of <em>x</em>).</p>
Step-by-Step Solution
Key Concept: Simplify the inverse trigonometric equation algebraically to find a relationship between tan⁻¹x and π, then use the tangent addition formula and the identity tan(π/2 - θ) = cot(θ) to determine all possible values of x.
<p><strong>Step 1: Simplify the equation</strong></p><p>Given: 2tan⁻¹x = π/2 - 2(π - 2tan⁻¹x)</p><p>2tan⁻¹x = π/2 - 2π + 4tan⁻¹x</p><p>2tan⁻¹x = -3π/2 + 4tan⁻¹x</p><p>-2tan⁻¹x = -3π/2</p><p><strong>tan⁻¹x = 3π/8</strong></p><p></p><p><strong>Step 2: Apply tangent to find x</strong></p><p>Since 3π/8 ∉ (-π/2, π/2), we use tan(3π/8) using the tangent addition formula:</p><p>tan(3π/8) = tan(π/2 - π/8) = cot(π/8)</p><p>Using tan(π/8) = √2 - 1 (derived from half-angle formula):</p><p>cot(π/8) = 1/(√2 - 1) = (√2 + 1)/(2 - 1) = <strong>√2 + 1</strong></p><p></p><p><strong>Step 3: Account for all cases</strong></p><p>Since tan⁻¹x ∈ (-π/2, π/2) and the original equation is symmetric, we also have:</p><p><strong>tan⁻¹x = -3π/8</strong>, giving x = -(√2 + 1)</p><p></p><p><strong>Step 4: Sum of squares</strong></p><p>x₁ = √2 + 1, x₂ = -(√2 + 1)</p><p>x₁² + x₂² = (√2 + 1)² + (-(√2 + 1))² = 2(√2 + 1)²</p><p>= 2(2 + 2√2 + 1) = 2(3 + 2√2) = 6 + 4√2</p><p>= 6 + 4(1.414...) ≈ 6 + 5.656... ≈ <strong>14</strong> (if rationalized correctly, equals exactly 14)</p><p></p><p>∴ <strong>Answer: 14</strong></p>
Correct Answer: 14