Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>If \(\tan\left[\tan^{-1}2 + \tan^{-1}20k\right] = k\), then the sum of all solutions \(k_1 + k_2\) equals?</p>
Step-by-Step Solution
Key Concept: Use the tangent addition formula tan(A + B) = (tan A + tan B)/(1 - tan A·tan B) with tan⁻¹2 and tan⁻¹(20k), then apply the constraint that the result equals k to form a quadratic equation.
<p><strong>Step 1:</strong> Apply tan(A + B) formula with A = tan⁻¹2, B = tan⁻¹(20k):</p><p>tan[tan⁻¹2 + tan⁻¹(20k)] = (2 + 20k)/(1 - 2·20k) = (2 + 20k)/(1 - 40k)</p><p><strong>Step 2:</strong> Set this equal to k:</p><p>(2 + 20k)/(1 - 40k) = k</p><p><strong>Step 3:</strong> Cross-multiply:</p><p>2 + 20k = k(1 - 40k)</p><p>2 + 20k = k - 40k²</p><p>40k² + 19k + 2 = 0</p><p><strong>Step 4:</strong> Use Vieta's formulas for sum of roots:</p><p>k₁ + k₂ = -19/40 = -0.475</p><p>∴ Answer: -0.475</p>
Correct Answer: -0.475