<p>Find the value of \(\tan^{-1}\left(\dfrac{1}{2}\tan 2A\right) + \tan^{-1}(\cot A) + \tan^{-1}(\cot^3 A)\) for \(0 < A < \dfrac{\pi}{4}\).</p>
Step-by-Step Solution
Key Concept: Convert each inverse tangent term using complementary angle relationships: cot A = tan(π/2 - A) and use tan(2A) = 2tan(A)/(1-tan²(A)). Then apply the addition formula for inverse tangents systematically.
<p><strong>Step 1:</strong> Rewrite cotangent terms. Note that cot A = tan(π/2 - A), so:</p><p>tan⁻¹(cot A) = tan⁻¹(tan(π/2 - A)) = π/2 - A (for 0 < A < π/2)</p><p><strong>Step 2:</strong> For the first term, use tan(2A) = 2tan(A)/(1-tan²(A)). Let t = tan A:</p><p>tan⁻¹(1/2 · tan(2A)) = tan⁻¹(t/(1-t²))</p><p>Note that d/dA[tan⁻¹(tan A)] = 1, and d/dA[tan A] = sec²(A), suggesting tan⁻¹(t/(1-t²)) relates to tan⁻¹(tan(2A)) via derivative patterns.</p><p><strong>Step 3:</strong> Recognize that tan⁻¹(1/2 · tan(2A)) = tan⁻¹(tan A) - tan⁻¹(tan³ A) (using composition of inverse tangent identities)</p><p>Therefore: tan⁻¹(1/2 · tan(2A)) = A - tan⁻¹(tan³ A)</p><p><strong>Step 4:</strong> Combine all three terms:</p><p>[A - tan⁻¹(tan³ A)] + [π/2 - A] + tan⁻¹(cot³ A)</p><p>= π/2 - tan⁻¹(tan³ A) + tan⁻¹(tan³(π/2 - A))</p><p>= π/2 - tan⁻¹(tan³ A) + tan⁻¹(cot³ A)</p><p><strong>Step 5:</strong> Use tan⁻¹(x) + tan⁻¹(1/x) = π/2 (for x > 0):</p><p>tan⁻¹(tan³ A) + tan⁻¹(cot³ A) = π/2</p><p>∴ Final answer: π/2 - (π/2) + (π/2) = <strong>π/2 radians</strong> or <strong>90°</strong></p><p><em>Note: If answer format requires degrees in range [0,180], the answer is <strong>90</strong></em></p>
Correct Answer: 180