Trigonometry & Inverse Trigonometry
General
Grade 12
Question:
<p>\(\tan^{-1}\!\left(1-x^2-\dfrac{1}{x^2}\right)+\sin^{-1}\!\left(x^2+\dfrac{1}{x^2}-1\right)\), \(x\ne 0\), equals:</p>
Step-by-Step Solution
<div class="solution"><p><strong>Key Idea:</strong> Let $t=x^2+1/x^2\ge 2$ (AM-GM).</p><p><strong>Step 1:</strong> Expression is $\tan^{-1}(1-t)+\sin^{-1}(t-1)$.</p><p><strong>Step 2:</strong> $\sin^{-1}(t-1)$ requires $t-1\in[-1,1]\implies t\in[0,2]$. Combined with $t\ge 2$: $t=2$ only.</p><p><strong>Step 3:</strong> At $t=2$: $\tan^{-1}(-1)+\sin^{-1}(1)=-\pi/4+\pi/2=\pi/4$</p><p><strong>Answer: (B) $\pi/4$</strong></p><div class="trap-box"><strong>Trap:</strong> Looking for a trig identity before checking domain -- the domain forces t=2.<div class="key-concept"><strong>Key Concept:</strong> Check domain constraints first; often they force a unique value
Correct Answer: 2