Q. If $y = \frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}},\$then$\frac{dy}{dx}\$is equal to
A) $\frac{1}{2}{cosec}^{2}\frac{x}{2}\$
B) $\frac{1}{2}cosec\frac{x}{2}$
C) $\frac{1}{2}{cosec}^{2}x$
D) ${cosec}^{2}\frac{x}{2}$
Explanation
(a)
$y = \frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}} \times \frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}$
$= \ \frac{2(1 + \cos{x)}}{- 2\sin x} = - \cot\frac{x}{2}$
$\therefore\frac{dy}{dx} = \frac{1}{2}{cosec}^{2}\frac{x}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.