Question:

<b>Simplify Half-Angle Then Differentiate</b><br> $\dfrac{1+\sin x}{1-\sin x} = \dfrac{1+\cos(\pi/2-x)}{1-\cos(\pi/2-x)}$. Let $\theta=\pi/2-x$:<br> $= \dfrac{1+\cos\theta}{1-\cos\theta} = \dfrac{2\cos^2(\theta/2)}{2\sin^2(\theta/2)} = \cot^2(\theta/2)$.<br> $y=\tan^{-1}(\cot(\theta/2))=\tan^{-1}\!\left(\tan\!\left(\dfrac{\pi}{2}-\dfrac{\theta}{2}\right)\right)=\dfrac{\pi}{2}-\dfrac{\theta}{2}=\dfrac{\pi}{2}-\dfrac{\pi/2-x}{2}=\dfrac{\pi}{4}+\dfrac{x}{2}$.<br> $\therefore\dfrac{dy}{dx}=\dfrac{1}{2}$. <b>Answer: 2 (= 1/2)</b><br> <b>Key concept:</b> Use $1\pm\sin x = 1\pm\cos(\pi/2-x)$ and double-angle formula to simplify inside $\tan^{-1}$.<br> <b>Trap:</b> Trying to differentiate $\tan^{-1}\sqrt{(1+\sin x)/(1-\sin x)}$ directly with chain rule — enormous computation versus the one-line simplification.