Q. $\frac{d}{dx}\sqrt{\frac{1 - \sin{2x}}{1 + \sin{2x}}\ }\$is equal to
A) $\sec^{2}x$
B) $- \sec^{2}\left( \frac{\pi}{4} - x \right)$
C) $\sec^{2}\left( \frac{\pi}{4} + x \right)$
D) $\sec^{2}\left( \frac{\pi}{4} - x \right)$
Explanation
(b)
Let $y = \sqrt{\frac{1 - \sin{2x}}{1 + \sin{2x}}} = \frac{\cos{x - \sin x}}{\cos{x + \sin x}}$
$= \frac{1 - \tan x}{1 + \tan x} = \tan\left( \frac{\pi}{4} - x \right)$
$\Rightarrow \ \$ $\frac{dy}{dx} = \ - \sec^{2}\left( \frac{\pi}{4} - x \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.