Question: If $\frac{\pi}{4}
Solution:
We have,
$\sqrt{1-\sin 2 x}$
$=\sqrt{\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x}$
$=\sqrt{(\sin x-\cos x)^{2}}$
$=|\sin x-\cos x|$
$=\sin x-\cos x$
$\left[\because \sin x>\cos x\right.$ for $\left.\frac{\pi}{4}
$\therefore \sqrt{1-\sin 2 x}=\sin x-\cos x$
