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