Question
If $\tan A=\frac{5}{6}, \tan B=\frac{1}{11}$ prove that $A+B=\frac{\pi}{4}$

Answer

Given $\tan A=\frac{5}{6}, \tan B=\frac{1}{11}$
$\begin{aligned}
& \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}=\frac{\frac{5}{6}+\frac{1}{11}}{1-\left(\frac{5}{6}\right)\left(\frac{1}{11}\right)} \\
&=\frac{\left(\frac{61}{66}\right)}{\left(\frac{61}{66}\right)}=1 \\
& \therefore \tan (A+B)=\tan \frac{\pi}{4} \\
& \therefore A+B=\frac{\pi}{4}
\end{aligned}$

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