Question

If $\sin A=\frac{1}{2},\cos B=\frac{12}{13}$, where $\frac{\mathrm{\pi }}{2}$< A < π and $\frac{3\mathrm{\pi }}{2}$ < B < 2π, find tan (A − B).

Answer 1

$$\begin{gathered} \text{Given:} \\ \sin A=\frac{1}{2}\mathrm{and}\cos B=\frac{12}{13} \\ \mathrm{Here},\frac{\mathrm{\pi }}{2}<A<\mathrm{\pi }\mathrm{and}\frac{3\mathrm{\pi }}{2}<B<2\mathrm{\pi }. \\ \text{That is,}A\mathrm{is}\mathrm{in}\text{the}\mathrm{second}\mathrm{quadrant}\mathrm{and}B\mathrm{is}\mathrm{in}\text{the}\mathrm{fourth}\mathrm{quadrant}. \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{in}\text{the}\mathrm{second}\mathrm{quadrant},\sin e\mathrm{function}\mathrm{is}\mathrm{positive}\mathrm{and}\cos \text{ine}\mathrm{and}\tan \mathrm{functions}\mathrm{are}\mathrm{negative}. \\ \mathrm{In}\text{the}\mathrm{fourth}\mathrm{quadran}t,\sin e\mathrm{and}\tan \mathrm{functions}\mathrm{are}\mathrm{negative}\mathrm{and}\cos ine\mathrm{function}\mathrm{is}\mathrm{positive}. \\ \mathrm{Therefore}, \\ \cos A=-\sqrt{1-{\sin }^{2}A}=-\sqrt{1-{\left(\frac{1}{2}\right)}^2}=-\sqrt{1-\frac{1}{4}}=-\sqrt{\frac{3}{4}}=\frac{-\sqrt{3}}{2} \\ \tan A\mathit{}=\frac{\sin A}{\cos A}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\displaystyle \raisebox{1ex}{$-\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{-1}{\sqrt{3}} \\ \sin B=-\sqrt{1-{\cos }^{2}B}=-\sqrt{1-{\left({\displaystyle \frac{12}{13}}\right)}^2}=-\sqrt{1-\frac{144}{169}}=-\sqrt{\frac{25}{169}}=\frac{-5}{13} \\ \tan B\mathit{}=\frac{\sin B}{\cos B}=\frac{-{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$13$}\right.}}{{\displaystyle \raisebox{1ex}{$12$}\!\left/ \!\raisebox{-1ex}{$13$}\right.}}=\frac{-5}{12} \\ \mathrm{Now},\tan \left(A-B\right)=\frac{\tan A-\tan B}{1+\tan A\tan B} \\ =\frac{{\displaystyle \frac{-1}{\sqrt{3}}}-{\displaystyle \frac{-5}{12}}}{1+{\displaystyle \frac{-1}{\sqrt{3}}}\times {\displaystyle \frac{-5}{12}}} \\ =\frac{{\displaystyle \frac{-12+5\sqrt{3}}{12\sqrt{3}}}}{{\displaystyle \frac{12\sqrt{3}+5}{12\sqrt{3}}}}=\frac{5\sqrt{3}-12}{5+12\sqrt{3}} \end{gathered}$$