Question

If $y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$, show that $\frac{dy}{dx}+{\sec }^2(\frac{\pi }{4}-x)$.

Answer 1

$y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$

$y=\sqrt{\frac{(\cos x-\sin x{)}^2}{(\cos x+\sin x{)}^2}}$

$y=\frac{1-\tan x}{1+\tan x}=\frac{\tan \pi /4-\tan x}{1+\tan \pi /4.\tan x}$

$y=\tan (\pi /4-x)$

$\frac{dy}{dx}=-{\sec }^2(\frac{\pi }{4}-x)$

$\frac{dy}{dx}+{\sec }^2(\pi /4-x)=0$