Question

$\frac{d}{dx}\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$ is equal to, $(0<x<\frac{\pi }{2})$,

• A. ${\sec }^{2}x$
• B. $-{\sec }^2(\frac{\pi }{4}-x)$
• C. ${\sec }^2(\frac{\pi }{4}+x)$
• D. ${\sec }^2(\frac{\pi }{4}-x)$

Answer 1

The correct option is B $-{\sec }^2(\frac{\pi }{4}-x)$

$y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$
$\sqrt{\frac{{\cos }^{2}x+{\sin }^{2}x-2\sin x\cos x}{{\cos }^{2}x+{\sin }^{2}x+2\sin x\cos x}}$
$\sqrt{\frac{{(\cos x-\sin x)}^2}{{(\cos x+\sin x)}^2}}$
$=\frac{\cos x-\sin x}{\cos x+\sin x}$
Dividing numerator & denominator by $\cos x$
$=\frac{1-\tan x}{1+\tan x}=\tan \left(\frac{\pi }{4}-x\right)$
Differentiate w r to $x$
$\therefore \frac{dy}{dx}=-{\sec }^2\left(\frac{\pi }{4}-x\right)$