Explanation

(c)
$\sin{4A - \cos{4A = \cos{2A - \sin{2A}}}}$
On squaring, we get
$1 - 2\sin{4\ A\cos{4\ A = 1 - 2\sin{2A\cos{2A}}}}$
$\Rightarrow \ \cos{4A = \frac{1}{2}}$
$\Rightarrow \ \tan{4A = \sqrt{3}}$
Alternate
Let $\tan{4A =}\sqrt{3} = \tan{\frac{\pi}{3}\ }$
$\Rightarrow \ \ A = \frac{\pi}{12}$
$\therefore\ \sin{4A - \cos{2A = \sin{\frac{\pi}{3} - \cos{\frac{\pi}{6} = 0}}}}$
And $\cos{4A - \sin{2A = \cos{\frac{\pi}{3} - \sin{\frac{\pi}{6} = 0}}}}$
$\therefore\ \sin{4A - \cos{2A = \cos{4A - \sin{2A}}}}$
Hence, our assumption is true.