#### Question

If \[\tan A = \frac{5}{6}\text{ and }\tan B = \frac{1}{11}\], prove that \[A + B = \frac{\pi}{4}\].

#### 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 .

Is there an error in this question or solution?

Solution If Tan a = 5 6 and Tan B = 1 11 , Prove that a + B = π 4 . Concept: Trigonometric Functions of Sum and Difference of Two Angles.