If $$(1 + \tan A) (1 + \tan B) = 2$$, then find $$A + B$$.

$$(1 + \tan A) (1 + \tan B) = 2$$

$$ 1 + \tan B + \tan A + \tan A \tan B = 2 $$

$$ \tan A + \tan B + \tan A \tan B = 1 $$

$$ \tan A + \tan B = 1 - \tan A \tan B $$

$$ \frac{ \tan A + \tan B}{1 - \tan A \tan B} = 1 $$

$$ \frac{ \tan x + \tan y}{1 - \tan x \tan y} = \tan (x+y) $$

$$ \frac{ \tan A + \tan B}{1 - \tan A \tan B} = \tan(A+B) = 1 $$

$$ \tan (A+B) = \tan 45^\circ $$

therefore,$$  A+B = 45^\circ $$