Trigonometry & Inverse Trigonometry
Compound Angles
Grade 11
Question:
<p>If <span class="math">\alpha + \beta = \frac{\pi}{2}\</span> and <span class="math">\beta + \gamma = \alpha\</span>, then the value of <span class="math">\tan\alpha\</span> is</p>
<p>(a) <span class="math">\tan\beta + \tan\gamma</span></p>
<p>(b) <span class="math">2(\tan\beta + \tan\gamma)</span></p>
<p>(c) <span class="math">\tan\beta + 2\tan\gamma</span></p>
<p>(d) <span class="math">2\tan\beta + \tan\gamma</span></p>
Step-by-Step Solution
Key Concept: Use the two given constraints to express β and γ in terms of α, then apply tangent addition formulas to find tan α. The key is recognizing that α = π/2 - β and β + γ = α allows us to eliminate variables.
<p><strong>Step 1:</strong> From the first constraint α + β = π/2, we get:<br>α = π/2 - β<br>Therefore: tan α = tan(π/2 - β) = cot β = 1/tan β</p><p><strong>Step 2:</strong> From the second constraint β + γ = α, substitute α = π/2 - β:<br>β + γ = π/2 - β<br>2β + γ = π/2<br>γ = π/2 - 2β</p><p><strong>Step 3:</strong> Find tan γ:<br>tan γ = tan(π/2 - 2β) = cot 2β = 1/tan 2β<br>Using tan 2β = 2tan β/(1 - tan²β), we get:<br>tan γ = (1 - tan²β)/(2tan β)</p><p><strong>Step 4:</strong> Calculate tan β + tan γ:<br>tan β + tan γ = tan β + (1 - tan²β)/(2tan β)<br>= (2tan²β + 1 - tan²β)/(2tan β)<br>= (tan²β + 1)/(2tan β)<br>= 1/(2tan β · cos²β) · cos²β = 1/(2sin β cos β) · cos²β</p><p><strong>Step 5:</strong> More directly, note that:<br>tan β + tan γ = tan β + (1 - tan²β)/(2tan β) = (2tan²β + 1 - tan²β)/(2tan β) = (1 + tan²β)/(2tan β)</p><p><strong>Step 6:</strong> Since tan α = 1/tan β = cot β, we have:<br>2(tan β + tan γ) = 2 · (1 + tan²β)/(2tan β) = (1 + tan²β)/tan β = 1/tan β = cot β = tan α</p><p><strong>∴ Answer:</strong> B</p>
Correct Answer: B