Trigonometry & Inverse Trigonometry
Trigonometric Equations
Grade 11
Question:
<p>If <span>\(\tan\alpha^2 = \tan(\alpha - \beta)\cdot\tan(\alpha + \beta)\)</span>, then which of the following is correct?</p><p>(Given: <span>\(0 < \alpha, \beta < 90^\circ\)</span>, <span>\(\tan\alpha > 0\)</span>)</p>
<p>(a) <span>\(\alpha = 30^\circ\)</span></p>
<p>(b) <span>\(\alpha = 45^\circ\)</span></p>
<p>(c) <span>\(\alpha = 60^\circ\)</span></p>
<p>(d) <span>\(\alpha = 90^\circ\)</span></p>
Step-by-Step Solution
Key Concept: Use the product-to-sum formula: tan(α−β)·tan(α+β) = [tan²α − tan²β]/[1 − tan²α·tan²β]. Since this equals tan²α, isolate tan²β to find the relationship between α and β.
<p><strong>Step 1:</strong> Apply the product formula for tan(α−β)·tan(α+β):</p><p>tan(α−β)·tan(α+β) = (tan α − tan β)/(1 + tan α tan β) · (tan α + tan β)/(1 − tan α tan β)</p><p>= (tan²α − tan²β)/(1 − tan²α tan²β)</p><p><strong>Step 2:</strong> Set this equal to tan²α:</p><p>tan²α − tan²β = tan²α(1 − tan²α tan²β)</p><p>tan²α − tan²β = tan²α − tan⁴α tan²β</p><p><strong>Step 3:</strong> Simplify:</p><p>tan²β = tan⁴α tan²β</p><p>tan²β(1 − tan⁴α) = 0</p><p>Since tan β ≠ 0 (from domain), we have: tan⁴α = 1</p><p><strong>Step 4:</strong> Since tan α > 0 and 0 < α < π/2, we get tan²α = 1, so tan α = 1</p><p>Therefore: <strong>α = π/4</strong></p><p>∴ Answer: B</p>
Correct Answer: B