Find out the value:

•(1+tan B)(1+tan A)=1+ (tan A+ tan B+ tan A. tan B) • tan (B+ A)= (tan A+ tan B)/(1-tan A. tan B) = tan A+ tan B+ tan A. tan B =tan 45°(given) = 1. so, 1+1= 2

If $A+B = 45^\circ$ , then we have:

$\begin{aligned} \tan (A+B) & = \tan 45^\circ \\ \frac {\tan A + \tan B}{1 - \tan A \tan B} & = 1 \\ \implies \tan A + \tan B & = 1 - \tan A\tan B \end{aligned}$

Now, we have:

$\begin{aligned} (1+\tan A) (1+\tan B) & = 1+{\color{#3D99F6}\tan A + \tan B} + \tan A \tan B & \small \color{#3D99F6} \text{Note that }\tan A + \tan B = 1 - \tan A\tan B \\ & = 1+{\color{#3D99F6}1 - \tan A\tan B} + \tan A \tan B \\ & = \boxed{2} \end{aligned}$