Question:
If $A$ and $B$ are acute angles such that $\tan A=\frac{1}{3}, \tan B=\frac{1}{2}$ and $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$, show that $A+B=45^{\circ}$.
Solution:
Given:
$\tan A=\frac{1}{3}$ and $\tan B=\frac{1}{2}$
$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$
On substituting these values in RHS of the expression, we get:
$\frac{\tan A+\tan B}{1-\tan A \tan B}=\frac{\left(\frac{1}{3}+\frac{1}{2}\right)}{\left(1-\frac{1}{3} \times \frac{1}{2}\right)}=\frac{\left(\frac{5}{6}\right)}{\left(1-\frac{1}{6}\right)}=\frac{\left(\frac{5}{6}\right)}{\left(\frac{5}{6}\right)}=1$
$\Rightarrow \tan (A+B)=1=\tan 45^{\circ} \quad\left[\because \tan 45^{\circ}=1\right]$
$\therefore \mathrm{A}+\mathrm{B}=45^{\circ}$