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Solution
$$A + B = 45^\circ$$
On taking tan both sides,
$$tan(A + B) = tan(45^\circ)$$
$$\frac{\tan A + \tan B}{1 - \tan A \tan B} = 1$$
$$\tan A + \tan B =Â 1 - \tan A \tan B$$
$$\tan A + \tan B + \tan A \tan BÂ = 1$$ ---(1)
Now,
$$2(1 + \tan A)(1 + \tan B)$$
= $$2(1 +Â \tan A +Â \tan B +Â \tan A \tan B)$$
from the eq(1),
= $$2(1 + 1)$$
= 4
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