Explanation

(b)
Since, $A + B + C = \pi$
$\Rightarrow a = \pi - (B + C)$
We have, $\cos{A = \cos{B\cos C}}$
$\Rightarrow \cos{\left\lbrack \pi - (B + C) \right\rbrack = \cos{B\cos C}}$
$\Rightarrow - \cos{(\ B + C) = \cos{B\cos C}}$
$\Rightarrow - \left\lbrack \cos{B\cos{C - \sin{B\sin C}}} \right\rbrack = \cos{B\cos C}$
$\Rightarrow \sin{B\ \sin\ C = 2\cos{B\cos C}}$
$\Rightarrow \tan{B\tan{C = 2}}$