Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p><strong>160.</strong> 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>(a) \(\dfrac{\sqrt{146}}{12}\)</p>
<p>(b) \(\dfrac{\sqrt{145}}{11}\)</p>
<p>(c) \(\dfrac{\sqrt{145}}{10}\)</p>
<p>(d) \(\dfrac{\sqrt{145}}{12}\)</p>
Step-by-Step Solution
Key Concept: Use the identity cos⁻¹(a) + cos⁻¹(b) = π/2 ⟺ cos⁻¹(a) = sin⁻¹(b), which means a = √(1-b²). This converts the equation into an algebraic relationship between the two arguments.
<p><strong>Step 1:</strong> Use the property that if cos⁻¹(a) + cos⁻¹(b) = π/2, then cos⁻¹(a) = sin⁻¹(b).</p><p>This means: <strong>a = cos(sin⁻¹(b)) = √(1-b²)</strong></p><p><strong>Step 2:</strong> Apply this with a = 2/(3x) and b = 3/(4x):</p><p>$$\frac{2}{3x} = \sqrt{1 - \left(\frac{3}{4x}\right)^2}$$</p><p><strong>Step 3:</strong> Square both sides:</p><p>$$\frac{4}{9x^2} = 1 - \frac{9}{16x^2}$$</p><p><strong>Step 4:</strong> Rearrange:</p><p>$$\frac{4}{9x^2} + \frac{9}{16x^2} = 1$$</p><p>$$\frac{64 + 81}{144x^2} = 1$$</p><p>$$\frac{145}{144x^2} = 1$$</p><p><strong>Step 5:</strong> Solve for x:</p><p>$$x^2 = \frac{145}{144}$$</p><p>$$x = \frac{\sqrt{145}}{12}$$ (taking positive root since x > 3/4)</p><p><strong>Step 6:</strong> Verify x > 3/4: Since √145 ≈ 12.04, we get x ≈ 1.003 > 0.75 ✓</p><p>∴ Answer: <strong>x = √145/12</strong></p>
Correct Answer: D