Question:

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