Question

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

Answer 1

$$\begin{gathered} \text{Given:} \\ \tan A=\frac{3}{4}\mathrm{and}\cos B=\frac{9}{41} \\ \mathrm{Here},\mathrm{\pi }<A<\frac{3\mathrm{\pi }}{2}\mathrm{and}0<B<\frac{\mathrm{\pi }}{2}. \\ \text{That is,}A\mathrm{is}\mathrm{in}\mathrm{third}\mathrm{quadrant}\mathrm{and}B\mathrm{is}\mathrm{in}\mathrm{first}\mathrm{qudrant}. \\ \mathrm{We}\mathrm{know}\mathrm{that}\tan \mathrm{function}\mathrm{is}\mathrm{positive}\mathrm{in}\mathrm{first}\mathrm{and}\mathrm{third}\mathrm{quadrant}s, \\ \mathrm{and}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{quadrant},\sin e\mathrm{function}\mathrm{is}\mathrm{also}\mathrm{positive}. \\ \mathrm{Therefore},\sin B=\sqrt{1-{\cos }^{2}B} \\ =\sqrt{1-{\left({\displaystyle \frac{9}{41}}\right)}^2} \\ =\sqrt{1-\frac{81}{1681}} \\ =\sqrt{\frac{1600}{1681}} \\ =\frac{40}{41} \\ \mathrm{And}\tan B=\frac{\sin B}{\cos B} \\ =\frac{{\displaystyle \raisebox{1ex}{$40$}\!\left/ \!\raisebox{-1ex}{$41$}\right.}}{{\displaystyle \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$41$}\right.}}=\frac{40}{9} \\ \mathrm{Therefore},\tan \left(A\mathit{+}B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B} \\ =\frac{{\displaystyle \frac{3}{4}}+{\displaystyle \frac{40}{9}}}{1-{\displaystyle \frac{3}{4}}\times {\displaystyle \frac{40}{9}}} \\ =\frac{{\displaystyle \frac{187}{36}}}{{\displaystyle \frac{-84}{36}}} \\ =\frac{-187}{84} \end{gathered}$$