Trigonometry & Inverse Trigonometry
Maximum and Minimum values of Trigonometric expressions
Grade 11
Question:
<p>If <i>A</i> > 0, <i>B</i> > 0 and <i>A</i> + <i>B</i> = π/6, then the maximum value of tan <i>A</i> + tan <i>B</i> is:</p>
<p>\(2\sqrt{3}\)</p>
<p>\(4 - 2\sqrt{3}\)</p>
<p>\(2 - \sqrt{3}\)</p>
<p>\(\sqrt{3} - 2\)</p>
Step-by-Step Solution
Key Concept: Use the tangent addition formula tan(A+B) = (tan A + tan B)/(1 - tan A·tan B) combined with the constraint A + B = π/6 to express tan A + tan B in terms of their product, then optimize.
<p><strong>Step 1:</strong> Apply the tangent addition formula with A + B = π/6:</p><p>tan(A + B) = tan(π/6) = 1/√3</p><p>Therefore: (tan A + tan B)/(1 - tan A·tan B) = 1/√3</p><p><strong>Step 2:</strong> Let tan A + tan B = S and tan A·tan B = P. Then:</p><p>S/(1 - P) = 1/√3</p><p>S = (1 - P)/√3</p><p><strong>Step 3:</strong> For tan A and tan B to be real and positive, they must be roots of t² - St + P = 0 with discriminant ≥ 0:</p><p>S² - 4P ≥ 0</p><p><strong>Step 4:</strong> Substitute S = (1 - P)/√3:</p><p>(1 - P)²/3 - 4P ≥ 0</p><p>(1 - P)² - 12P ≥ 0</p><p>1 - 2P + P² - 12P ≥ 0</p><p>P² - 14P + 1 ≥ 0</p><p><strong>Step 5:</strong> This gives P ≤ 7 - 4√3 (taking the relevant root for small P). S is maximized when P is minimized at P = 7 - 4√3:</p><p>S = (1 - (7 - 4√3))/√3 = (4√3 - 6)/√3 = 4 - 2√3</p><p>∴ Answer: B</p>
Correct Answer: B