Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>A number \(k\) is such that \(\tan[\arctan(2) + \arctan(20k)] = k\). The sum of all possible values of \(k\) is ______.</p>
Step-by-Step Solution
Key Concept: Use the tangent addition formula tan(A+B) = (tan A + tan B)/(1 - tan A·tan B) where A = arctan(2) and B = arctan(20k), then set the result equal to k and solve the resulting equation.
<p><strong>Step 1:</strong> Apply the tangent addition formula. Let tan(arctan(2)) = 2 and tan(arctan(20k)) = 20k.</p><p>tan[arctan(2) + arctan(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 (valid when 1 - 40k ≠ 0):</p><p>2 + 20k = k(1 - 40k)</p><p>2 + 20k = k - 40k²</p><p>40k² + 20k - k + 2 = 0</p><p>40k² + 19k + 2 = 0</p><p><strong>Step 4:</strong> Use the quadratic formula:</p><p>k = (-19 ± √(361 - 320))/80 = (-19 ± √41)/80</p><p><strong>Step 5:</strong> Verify both solutions satisfy 1 - 40k ≠ 0 (i.e., k ≠ 1/40). Both values clearly do not equal 1/40.</p><p><strong>Step 6:</strong> Find the sum of roots using Vieta's formula:</p><p>Sum = -19/40</p><p>∴ Answer: <strong>-19/40</strong></p>
Correct Answer: -19