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Question
If \[\tan A = \frac{5}{6}\text{ and }\tan B = \frac{1}{11}\], prove that \[A + B = \frac{\pi}{4}\].
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Solution
We have:
\[\tan A = \frac{5}{6}\text{ and }\tan B = \frac{1}{11}\]
\[\text{ Therefore, }\tan\left( A + B \right) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\]
\[ \Rightarrow \tan\left( A + B \right) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\]
\[ \Rightarrow \tan\left( A + B \right) = \frac{\frac{5}{6} + \frac{1}{11}}{1 - \frac{5}{6} \times \frac{1}{11}}\]
\[ \Rightarrow \tan\left( A + B \right) = \frac{\frac{61}{66}}{\frac{61}{66}}\]
\[ \Rightarrow \tan\left( A + B \right) = 1\]
\[ \Rightarrow \tan\left( A + B \right) = \tan\left( \frac{\pi}{4} \right)\]
\[\text{ Therefore, }A + B = \frac{\pi}{4} . \]
Hence proved .
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