Trigonometry & Inverse Trigonometry
General
Grade 12

Question:

<p>If <span class="math-inline">\(\alpha=2\tan^{-1}\!\frac{1+x}{1-x}\)</span> and <span class="math-inline">\(\beta=\sin^{-1}\!\frac{1-x^2}{1+x^2}\)</span> for <span class="math-inline">\(x>1\)</span>, then <span class="math-inline">\(\alpha+\beta=\)</span></p>
<strong>-π</strong>
π
π/2

Step-by-Step Solution

Key Concept: General
<div class="solution"><p><strong>Step 1:</strong> Let <span class="math-inline">$x=\tan\theta$</span>, <span class="math-inline">$x>1\implies\theta\in(\pi/4,\pi/2)$</span>.</p><p><strong>Step 2:</strong> <span class="math-inline">$\beta=\sin^{-1}(\cos 2\theta)=\sin^{-1}(\sin(\pi/2-2\theta))$</span>. Since <span class="math-inline">$\pi/2-2\theta\in(-\pi/2,0)$</span>: <span class="math-inline">$\beta=\pi/2-2\theta$</span>.</p><p><strong>Step 3:</strong> <span class="math-inline">$\frac{1+x}{1-x}=\frac{1+\tan\theta}{1-\tan\theta}=\tan(\pi/4+\theta)$</span>. Since <span class="math-inline">$\pi/4+\theta>\pi/2$</span>: <span class="math-inline">$\alpha=2(\pi/4+\theta-\pi)=2\theta-3\pi/2$</span>.</p><p><strong>Step 4:</strong> <span class="math-inline">$\alpha+\beta=(2\theta-3\pi/2)+(\pi/2-2\theta)=-\pi$</span>.</p><p><strong>Answer: (A) <span class="math-inline">$-\pi$</span></strong></p><div class="trap-box"><strong>Trap:</strong> Not subtracting π when tan⁻¹ argument exceeds the principal range.</div><div class="key-concept"><strong>Key Concept:</strong> Branch tracking is mandatory when argument of tan⁻¹ moves outside principal interval</div></div>
Correct Answer: 1

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