Explanation

(a)
$\because\sin{A\sin{B = \frac{1}{2} \times \ 2\sin{A\sin B}}}$
$= \frac{1}{2}\left\lbrack \cos{(\ A - B) - \cos{(A + B)}} \right\rbrack$
$= \ \frac{1}{2}\left\lbrack \cos{(\ A - B) - \cos{90{^\circ}}} \right\rbrack$ $(\because A + B + C = 180{^\circ}\ and\ \angle C = 90{^\circ},\$given)
$= \frac{1}{2}\cos{(A - B) \leq \ \frac{1}{2}}$
$\therefore$ Maximum value of $\sin{A\sin{B = \frac{1}{2}}}$