Explanation

(c)
Given, $f(x) = \frac{2\sin{8x\cos{x - 2\sin{6x\cos{3x}}}}}{2\cos{2x\cos{x - 2\sin{3x\sin{4x}}}}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{\left( \sin{9x + \sin{7x}} \right) + (\sin{9x + \sin{3x}})}{\left( \cos{3x + \cos x} \right) + (\cos{7x - \cos x})}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{\sin{7x - \sin{3x}}}{\cos{7x + \cos{3x}}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{2\cos{5x\sin{2x}}}{2\cos{2x\cos{5x}}} = \tan{2x}$
$\therefore$ Period of $f(x) = \frac{\pi}{2}$