Question

If \[\tan A = \frac{3}{4}, \cos B = \frac{9}{41}\], where π < A < \[\frac{3\pi}{2}\] and 0 < B <\[\frac{\pi}{2}\], find tan (A + B).

 

Answer 1

Given:
\[\tan A = \frac{3}{4}\text{ and }\cos B = \frac{9}{41}\]
\[\text{ Here,} \pi < A < \frac{3\pi}{2}\text{ and }0 < B < \frac{\pi}{2} . \]
That is, A is in third quadrant and B is in first qudrant . 
We know that tan function is positive in first and third quadrants, and in the first quadrant, \sine function is also positive . 
\[\text{ Therefore, }\sin B = \sqrt{1 - \cos^2 B}\]
\[ = \sqrt{1 - \left( \frac{9}{41} \right)^2}\]
\[ = \sqrt{1 - \frac{81}{1681}}\]
\[ = \sqrt{\frac{1600}{1681}}\]
\[ = \frac{40}{41}\]
\[\text{ And }\tan B = \frac{\sin B}{\cos B}\]
\[ = \frac{\frac{40}{41}}{\frac{9}{41}} = \frac{40}{9}\]
\[\text{Therefore, }\tan\left( A + B \right) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\]
\[ = \frac{\frac{3}{4} + \frac{40}{9}}{1 - \frac{3}{4} \times \frac{40}{9}}\]
\[ = \frac{\frac{187}{36}}{\frac{- 84}{36}}\]
\[ = \frac{- 187}{84}\]