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Atan A + tan B = 0

Btan A = tan B

C $\frac{2tan A+ tan B}{1-tan A tan B}=1$

D $\frac{2tan A+ tan B}{1+ tan A tan B}=0$

Answer:

$tan (A+B) = \frac{tan A + tan B}{1-tanAtanB}$
so tan (A+B)=0 
the numerator becomes zero

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