Question:

<div class="solution"><p><strong>Key Idea:</strong> Use half-angle: \(1\pm\sin x=(\sin(x/2)\pm\cos(x/2))^2\).</p><p><strong>Step 1:</strong> For \(\pi/2<x<\pi\): \(x/2\in(\pi/4,\pi/2)\), so \(\sin(x/2)>\cos(x/2)>0\).</p><p><strong>Step 2:</strong> <span class="math-block">\[\sqrt{1-\sin x}=\sin(x/2)-\cos(x/2),\quad\sqrt{1+\sin x}=\sin(x/2)+\cos(x/2)\]</p><p><strong>Step 3:</strong> Numerator \(=2\sin(x/2)\), Denominator \(=-2\cos(x/2)\). Ratio \(=-\tan(x/2)\).</p><p><strong>Step 4:</strong> \(\cot^{-1}(-\tan(x/2))=\pi-\cot^{-1}(\tan(x/2))=\pi-(\pi/2-x/2)=\pi/2+x/2\)</p><p><strong>Answer: (B) \(\pi/2+x/2\)</strong></p><div class="trap-box"><strong>Trap:</strong> Getting the sign of \(\sin(x/2)-\cos(x/2)\) wrong leads to option (C).<div class="key-concept"><strong>Key Concept:</strong> Half-angle squares + quadrant sign check before simplifying radicals