If A and B are acute angles such that tan A


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 $A+B=45^{\circ}$.




$\tan A=\frac{1}{3}$ 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:

$\frac{\tan A+\tan B}{1-\tan A \tan B}=\frac{\left(\frac{1}{3}+\frac{1}{2}\right)}{\left(1-\frac{1}{3} \times \frac{1}{2}\right)}=\frac{\left(\frac{5}{6}\right)}{\left(1-\frac{1}{6}\right)}=\frac{\left(\frac{5}{6}\right)}{\left(\frac{5}{6}\right)}=1$

$\Rightarrow \tan (A+B)=1=\tan 45^{\circ} \quad\left[\because \tan 45^{\circ}=1\right]$

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


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