Trigonometry & Inverse Trigonometry
General
Grade 12

Question:

<p>\(\text{cosec}^{-1}(\cos x)\) exists if:</p>
x\in [-1,1]
x\in R
x is an odd multiple of \pi/2
x is a multiple of \pi

Step-by-Step Solution

<div class="solution"><p><strong>Key Idea:</strong> Domain of $\text{cosec}^{-1}(u)$ requires $|u|\ge 1$.</p><p><strong>Step 1:</strong> Need $|\cos x|\ge 1$. Since $\cos x\in[-1,1]$, we need $|\cos x|=1$.</p><p><strong>Step 2:</strong> $\cos x=\pm 1$ iff $x=n\pi$.</p><p><strong>Answer: (D) x is a multiple of \pi</strong></p><div class="trap-box"><strong>Trap:</strong> Confusing with odd multiples of \pi/2 (where cos x = 0, not \pm1).<div class="key-concept"><strong>Key Concept:</strong> cosec⁻^1 and sec⁻^1 need |argument| \ge 1 -- only attained at extremes of cos/sin
Correct Answer: 4

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