Answer
See the steps.
Work Step by Step
$\sin{(90+x)} = \sin{90} \cos{x} + \sin{x} \cos{90} $
$\sin{(90+x)} = 1 \times \cos{x} + \sin{x} \times 0 = \cos{x}$
$\sin{(90-x)} = \sin{90} \cos{x} - \sin{x} \cos{90} $
$\sin{(90-x)} = 1 \times \cos{x} - \sin{x} \times 0 = \cos{x}$
$LHS = \sin{(90+x)} -\sin{(90-x)} = \cos{x} - \cos{x} = 0 $
$\therefore LHS = RHS$