If \(y = \log_e\sqrt{\displaystyle\frac{1 - \cos x}{1 + \cos x}}\), where \(0 \lt x \lt \pi\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\mathrm{cosec}\, x\)
By the half-angle formulas, \(1 - \cos x = 2\sin^{2}\frac{x}{2}\) and \(1 + \cos x = 2\cos^{2}\frac{x}{2}\):
\( \displaystyle \sqrt{\frac{1 - \cos x}{1 + \cos x}} = \tan\frac{x}{2} \)
(positive, since \(0 \lt \frac{x}{2} \lt \frac{\pi}{2}\)). So \(y = \log_e\tan\frac{x}{2}\), and by the chain rule:
\( \displaystyle \frac{dy}{dx} = \frac{1}{\tan\frac{x}{2}} \cdot \sec^{2}\frac{x}{2} \cdot \frac{1}{2} \)
\( \displaystyle = \frac{1}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \frac{1}{\sin x} = \mathrm{cosec}\, x \)