If $y = \sqrt{\dfrac{1-\sin 2x}{1+\sin 2x}}$ show that $\dfrac{dy}{dx}+\sec^2 \left(\dfrac{\pi}{4}-x\right)=0$
$
\begin{aligned}
\dfrac{1-\sin 2x}{1+\sin 2x} & = \dfrac{\sin^2 x + \cos^2 x -2 \sin x \cos x}{\sin^2 x + \cos^2 x + 2 \sin x \cos x} \\
& = \dfrac{\left(\cos x - \sin x\right)^2}{\left(\cos x + \sin x\right)^2} \\
\therefore y = \sqrt{\dfrac{1-\sin 2x}{1+\sin 2x}} & = \dfrac{\cos x - \sin x}{\cos x + \sin x} \\
& = \dfrac{\left(\cos x - \sin x\right)^2}{\left(\cos x + \sin x\right) \left(\cos x - \sin x\right)} \\
& = \dfrac{\cos^2 x + \sin^2 x -2 \sin x \cos x}{\cos^2 x - \sin^2 x} \\
& = \dfrac{1-\sin 2x}{\cos 2x}
\end{aligned}
$
$
\begin{aligned}
\therefore \dfrac{dy}{dx} & = \dfrac{\cos 2x \times \dfrac{d}{dx}\left(1-\sin 2x\right)-\left(1-\sin 2x\right)\times \dfrac{d}{dx}\cos 2x}{\cos^2 2x} \\
& = \dfrac{\cos 2x \times \left(-2 \cos 2x\right) - \left(1-\sin 2x\right) \times \left(-2 \sin 2x\right)}{\cos^2 2x} \\
& = \dfrac{-2 \cos^2 2x - 2 \sin^2 2x + 2 \sin 2x}{\cos^2 2x} \\
& = \dfrac{-2\left(\cos^2 2x + \sin^2 2x\right) + 2 \sin 2x}{\cos^2 2x} \\
& = \dfrac{-2 + 2 \sin 2x}{\cos^2 2x} \\
& = \dfrac{-2 \left(1-\sin 2x\right)}{1-\sin^2 2x} \\
& = \dfrac{-2 \left(1-\sin 2x\right)}{\left(1+\sin 2x\right)\left(1-\sin 2x\right)} \\
& = \dfrac{-2}{1+\sin 2x}
\end{aligned}
$
$
\begin{aligned}
\text{Now, } \cos \left(\dfrac{\pi}{4}-x\right) & = \cos \left(\dfrac{\pi}{4}\right) \cos x + \sin \left(\dfrac{\pi}{4}\right) \sin x \\
& = \dfrac{1}{\sqrt{2}} \cos x + \dfrac{1}{\sqrt{2}} \sin x \\
& = \dfrac{1}{\sqrt{2}} \left(\cos x + \sin x\right) \\
\therefore \cos^2 \left(\dfrac{\pi}{4}-x\right) & = \dfrac{1}{2} \left(\cos x + \sin x\right)^2 \\
& = \dfrac{1}{2} \left(\cos^2 x + \sin^2 x +2 \sin x \cos x\right) \\
& = \dfrac{1+\sin 2x}{2} \\
\therefore \sec^2 \left(\dfrac{\pi}{4}-x\right) & = \dfrac{2}{1+\sin 2x}
\end{aligned}
$
$\therefore$ $\dfrac{dy}{dx} = \dfrac{-2}{1+\sin 2x} = - \sec^2 \left(\dfrac{\pi}{4}-x\right)$
i.e. $\dfrac{dy}{dx} + \sec^2 \left(\dfrac{\pi}{4} - x\right) = 0$