Question

If $y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$. Prove that $\frac{dy}{dx}+{\sec }^2(\frac{\pi }{4}-x)=0$

Answer 1

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

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

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

$=\frac{|\sin x-\cos x|}{\sin x+\cos x}=\frac{\cos x-\sin x}{\cos x+\sin x}$

$\frac{dy}{dx}=\left[\frac{(\cos x+\sin x)(-\sin x-\cos x)-(\cos x-\sin x)(-\cos x+\sin x)}{(\sin x+\cos x{)}^2}\right]$

$=-\left[\frac{(\cos x+\sin x{)}^2+(\sin x-\cos x{)}^2}{(\sin x+\cos x{)}^2}\right]$

$=-\left[\frac{{\cos }^{2}x+{\sin }^{2}x+{\cos }^{2}x+{\sin }^{2}x}{(\sin x+\cos x{)}^2}\right]$

$=\frac{-2}{(\sin x+\cos x{)}^2}=\frac{-1}{(\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x{)}^2}$

$\frac{dy}{dx}=\frac{-1}{\left(\cos \right(\pi /4-x){)}^2}=-se{c}^2(\frac{\pi }{4}-x)$

$\Rightarrow \frac{dy}{dx}=se{c}^2(\frac{\pi }{4}-x)=0$