Question

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

Answer 1

Given, A and B are acute angles such that $\tan A=\frac{1}{2}\tan B=\frac{1}{3}$

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

$\tan (A+B)=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\frac{1}{3}}$

$\tan (A+B)=\frac{\frac{5}{6}}{1-\frac{1}{6}}$

$\tan (A+B)=\frac{\frac{5}{6}}{\frac{5}{6}}$

$\tan (A+B)=1$

we know, $\tan {45}^{\circ }=1$

$\therefore (A+B)={45}^{\circ }$