Answer
$a$
Work Step by Step
$\sin{\alpha} + \sin{\beta} =2 \sin{(\dfrac{\alpha + \beta}{2})} \cos{(\dfrac{\alpha -\beta}{2})}$
$\sin{4x} + \sin{2x} = 2 \sin{3x} \cos{x}$
$\cos{\alpha} -\cos{\beta} = -2 \sin{(\dfrac{\alpha+\beta}{2})} \sin{(\dfrac{\alpha-\beta}{2})} $
$\cos{4x} - \cos{2x} = -2 \sin{3x} \sin{x}$
$LHS = \dfrac{2 \sin{3x} \cos{x}}{-2 \sin{3x} \sin{x}}= -\cot{x} = (a)$
