Question

If $\sin A=\frac{1}{2},\cos B=\frac{\sqrt{3}}{2}$, where $\frac{\mathrm{\pi }}{2}$ < A < π and 0 < B < $\frac{\mathrm{\pi }}{2}$, find the following:

(i) tan (A + B)
(ii) tan (A − B)

Answer 1

$$\begin{gathered} \text{Given:}\sin A=\frac{1}{2}\mathrm{and}\cos B=\frac{\sqrt{3}}{2} \\ \mathrm{Here},\frac{\mathrm{\pi }}{2}<A<\mathrm{\pi }\mathrm{and}0<B<\frac{\mathrm{\pi }}{2}. \\ \text{That is,}A\mathrm{is}\mathrm{in}\text{the}\mathrm{second}\mathrm{quadrant}\mathrm{and}B\mathrm{is}\mathrm{in}\text{the}\mathrm{first}\mathrm{quadrant}. \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{in}\text{the}\mathrm{second}\mathrm{quadrant},\sin \text{e}\mathrm{function}\mathrm{is}\mathrm{positive}\mathrm{and}\cos \text{ine}\mathrm{and}\tan \mathrm{functions}\mathrm{are}\mathrm{negative} \\ \mathrm{In}\text{the}\mathrm{first}\mathrm{quadrant},\mathrm{all}\mathrm{T}-\mathrm{functions}\mathrm{are}\mathrm{positive}. \\ \mathrm{Therefore}, \\ \cos A=-\sqrt{1-{\sin }^{2}A}=-\sqrt{1-{\left(\frac{1}{2}\right)}^2}=-\sqrt{1-\frac{1}{4}}=-\sqrt{\frac{3}{4}}=\frac{-\sqrt{3}}{2} \\ \tan A=\frac{\sin A}{\cos A}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\displaystyle \raisebox{1ex}{$-\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{-1}{\sqrt{3}} \\ \sin B\mathit{}=\sqrt{1-{\cos }^{2}A}=\sqrt{1-{\left(\frac{\sqrt{3}}{2}\right)}^2}=\sqrt{1-\frac{3}{4}}=\sqrt{\frac{1}{4}}=\frac{1}{2} \\ \tan B\mathit{}=\frac{\sin B}{\cos B}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\displaystyle \raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{1}{\sqrt{3}} \end{gathered}$$

$$\begin{gathered} \mathrm{Now}, \\ \left(\mathrm{i}\right)\tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B} \\ =\frac{{\displaystyle \frac{-1}{\sqrt{3}}}+{\displaystyle \frac{1}{\sqrt{3}}}}{1-{\displaystyle \frac{-1}{\sqrt{3}}}\times {\displaystyle \frac{1}{\sqrt{3}}}} \\ =\frac{0}{1+{\displaystyle \frac{1}{3}}}=0 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\tan \left(A-B\right)=\frac{\tan A-\tan B}{1+\tan A\tan B} \\ =\frac{{\displaystyle \frac{-1}{\sqrt{3}}}-{\displaystyle \frac{1}{\sqrt{3}}}}{1+{\displaystyle \frac{-1}{\sqrt{3}}}\times {\displaystyle \frac{1}{\sqrt{3}}}} \\ =\frac{{\displaystyle \frac{-2}{\sqrt{3}}}}{1-{\displaystyle \frac{1}{3}}} \\ =\frac{{\displaystyle \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$\sqrt{3}$}\right.}}{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}} \\ =-\sqrt{3} \end{gathered}$$