(d) We have $\cos A = \cos B\cos C$ By triangle property,

$A + B + C = \pi \Rightarrow B + C = \pi – A$

$\therefore \cos (B + C) = \cos (\pi – A) \Rightarrow \cos (B + C) = – \cos A$

$ \Rightarrow \cos B\cos C – \sin B\sin C = – \cos B\cos C$

$( \because {\rm{Given}}\cos A = \cos B\cos C)$

$ \Rightarrow 2\cos B\cos C = \sin B\sin C$

$ \Rightarrow \frac{{\cos B\cos C}}{{\sin B\sin C}} = \frac{1}{2}$

$\Rightarrow \cot B\cot C = \frac{1}{2}$.