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If $\tan A=\frac{5}{6}, \tan B=\frac{1}{11}$, then $A+B=$
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The correct answer is:
$\frac{\pi}{4}$
$\begin{aligned} \tan (A+B) &=\frac{\tan A+\tan B}{1-\tan A \tan B} \\ &=\frac{\frac{5}{6}+\frac{1}{11}}{1-\frac{5}{6} \times \frac{1}{11}}=\frac{\frac{55+6}{66}}{1-\frac{5}{66}}=\frac{\left(\frac{61}{66}\right)}{\left(\frac{61}{66}\right)} \\ \tan (A+B) &=1 \Rightarrow A+B=\tan ^{-1}(1) \Rightarrow A+B=\frac{\pi}{4} \end{aligned}$
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