Trigonometry & Inverse Trigonometry
General
Grade 12
Question:
<p>If <span class="math-inline">\(\sin^{-1}\!\frac{2\alpha}{1+\alpha^2}+\sin^{-1}\!\frac{2\beta}{1+\beta^2}=2\tan^{-1}x\)</span>, then <span class="math-inline">\(x=\)</span></p>
<strong>(α+β)/(1-αβ)</strong>
α+β
α-β
(α-β)/(1+αβ)
Step-by-Step Solution
Key Concept: General
<div class="solution"><p>Use <span class="math-inline">\(\sin^{-1}\!\frac{2t}{1+t^2}=2\tan^{-1}t\)</span>. So equation becomes <span class="math-inline">\(2\tan^{-1}\alpha+2\tan^{-1}\beta=2\tan^{-1}x\implies x=\frac{\alpha+\beta}{1-\alpha\beta}\)</span>.</p><p><strong>Answer: (1) <span class="math-inline">\(\frac{\alpha+\beta}{1-\alpha\beta}\)</span></strong></p><div class="key-concept"><strong>Key Concept:</strong> When <span class="math-inline">\(\frac{2t}{1+t^2}\)</span> appears inside sin⁻¹, substitute <span class="math-inline">\(2\tan^{-1}t\)</span></div></div>
Correct Answer: 1