Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>If \( \alpha = \cos^{-1}\!\left(\dfrac{3}{5}\right) \), \( \beta = \tan^{-1}\!\left(\dfrac{1}{3}\right) \), where \( 0 < \alpha,\, \beta < \dfrac{\pi}{2} \), then \( \alpha - \beta \) is equal to:</p>
<p>\( \tan^{-1}\!\left(\dfrac{9}{5\sqrt{10}}\right) \)</p>
<p>\( \cos^{-1}\!\left(\dfrac{9}{5\sqrt{10}}\right) \)</p>
<p>\( \tan^{-1}\!\left(\dfrac{9}{14}\right) \)</p>
<p>\( \sin^{-1}\!\left(\dfrac{9}{5\sqrt{10}}\right) \)</p>
Step-by-Step Solution
Key Concept: Convert inverse trigonometric functions to their corresponding triangle representations, then use the tangent addition formula. Since α and β are in (0, π/2), we can construct right triangles to find tan(α) and tan(β) explicitly.
<p><strong>Step 1:</strong> From α = cos⁻¹(3/5) where 0 < α < π/2, construct a right triangle with adjacent = 3, hypotenuse = 5. Then opposite = √(25 - 9) = 4.</p><p>Therefore, tan(α) = 4/3.</p><p><strong>Step 2:</strong> Given β = tan⁻¹(1/3) where 0 < β < π/2, we have tan(β) = 1/3.</p><p><strong>Step 3:</strong> Use the tangent addition formula:</p><p>tan(α + β) = (tan α + tan β)/(1 - tan α · tan β)</p><p>tan(α + β) = (4/3 + 1/3)/(1 - 4/3 · 1/3)</p><p>tan(α + β) = (5/3)/(1 - 4/9) = (5/3)/(5/9) = (5/3) × (9/5) = 3</p><p><strong>Step 4:</strong> Since 0 < α + β < π and tan(α + β) = 3 > 0, we have α + β ∈ (0, π/2).</p><p>Therefore, α + β = tan⁻¹(3).</p><p>∴ Answer: A</p>
Correct Answer: A