Question

If A and B are acute angles such that tan A = 1/2,  tan B = 1/3 and tan (A + B) = ${\displaystyle \frac{\tan A+\tan B}{1-\tan A\tan B}}$ A + B = ?

Answer 1

${\displaystyle \tan A=\frac{1}{2}, \tan B=\frac{1}{3}}$

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

${\displaystyle \mathrm{Tan}(A+B)=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2.1}/3}=\frac{\frac{5}{6}}{1-\frac{1}{6}}=1}$

Tan (A + B) = Tan 45°

∴ A + B = 45°