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(a) Find the values of $A$ and $B$ ($0 \leq A < 90^{\circ}, 0 \leq B < 90^{\circ}$), if $\tan(A + B) = 1$ and $\tan(A - B) = \frac{1}{\sqrt{3}}$.
OR
(b) Prove that $\tan 45^{\circ} = 1$ geometrically.
OR
(b) Prove that $\tan 45^{\circ} = 1$ geometrically.
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Solution: (a) $A + B = 45^{\circ}$
$A - B = 30^{\circ}$
Solving and getting $A = 37.5^{\circ}$ and $B = 7.5^{\circ}$
OR
(b) Consider an isosceles right $\Delta ABC$
Using angle sum property $\angle A = \angle C = 45^{\circ}$
Clearly, $\tan 45^{\circ} = \frac{AB}{BC} = \frac{x}{x} = 1$
$A - B = 30^{\circ}$
Solving and getting $A = 37.5^{\circ}$ and $B = 7.5^{\circ}$
OR
(b) Consider an isosceles right $\Delta ABC$
Using angle sum property $\angle A = \angle C = 45^{\circ}$
Clearly, $\tan 45^{\circ} = \frac{AB}{BC} = \frac{x}{x} = 1$