Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>If \( \cos^{-1}\!\left(\dfrac{2}{3x}\right) + \cos^{-1}\!\left(\dfrac{3}{4x}\right) = \dfrac{\pi}{2} \left(x > \dfrac{3}{4}\right) \), then \( x \) is equal to</p>
<p>\( \dfrac{\sqrt{145}}{12} \)</p>
<p>\( \dfrac{\sqrt{145}}{10} \)</p>
<p>\( \dfrac{\sqrt{146}}{12} \)</p>
<p>\( \dfrac{\sqrt{145}}{11} \)</p>
Step-by-Step Solution
Key Concept: When cos⁻¹(a) + cos⁻¹(b) = π/2, then cos⁻¹(a) = sin⁻¹(b), which means a = √(1-b²). This converts the inverse trigonometric equation into an algebraic equation.
<p><strong>Step 1:</strong> Use the identity: If cos⁻¹(a) + cos⁻¹(b) = π/2, then cos⁻¹(a) = π/2 - cos⁻¹(b) = sin⁻¹(b)</p><p>Therefore: <span style='font-family:monospace;'>2/(3x) = sin(cos⁻¹(3/(4x)))</span></p><p><strong>Step 2:</strong> If cos(θ) = 3/(4x), then sin(θ) = √(1 - 9/(16x²)) = √(16x² - 9)/(4x)</p><p>So: <span style='font-family:monospace;'>2/(3x) = √(16x² - 9)/(4x)</span></p><p><strong>Step 3:</strong> Cross-multiply: <span style='font-family:monospace;'>8x = 3x√(16x² - 9)</span></p><p>Simplify: <span style='font-family:monospace;'>8 = 3√(16x² - 9)</span></p><p><strong>Step 4:</strong> Square both sides: <span style='font-family:monospace;'>64 = 9(16x² - 9)</span></p><p><span style='font-family:monospace;'>64 = 144x² - 81</span></p><p><span style='font-family:monospace;'>145 = 144x²</span></p><p><span style='font-family:monospace;'>x² = 145/144</span></p><p><span style='font-family:monospace;'>x = √145/12</span> (taking positive root since x > 3/4)</p><p><strong>Verification:</strong> √145/12 ≈ 1.006 > 3/4 ✓, and both 2/(3x) and 3/(4x) are ≤ 1 ✓</p><p>∴ Answer: <strong>x = √145/12</strong></p>
Correct Answer: A