Explanation

(d)
$\tan{A\tan{B = 2}}$
$\Rightarrow \ \ \frac{\sin{A\sin B}}{\cos{Acos\ B}} = 2$
Using componendo and dividendo, we get
$\frac{\sin{A\sin B + \cos{A\cos B}}}{\sin{A\sin B - \cos{A\cos B}}} = \frac{2 + 1}{2 - 1}$
$\Rightarrow \ \ \frac{\cos{(A - B)}}{- \cos{(A + B)}} = \frac{3}{1}$
$\Rightarrow \ \ \frac{3/5}{- \cos{(A + B)}} = \frac{3}{1}\ \ \ \ \ \ \ \left\lbrack \because\cos{(A - B) = \frac{3}{5}},\ given \right\rbrack$
$\Rightarrow \ \cos{(A + B) = - \frac{1}{5}}$