Question

If $\cos A=-\frac{12}{13}\mathrm{and}\cot B=\frac{24}{7}$, where A lies in the second quadrant and B in the third quadrant, find the values of the following:

(i) sin (A + B)
(ii) cos (A + B)
(iii) tan (A + B)

Answer 1

$$\begin{gathered} \text{Given:} \\ \cos A=-\frac{12}{13}\mathrm{and}\cot B=\frac{24}{7} \\ A\mathrm{lies}\mathrm{in}\text{the}\mathrm{second}\mathrm{quadrant}\mathrm{and}B\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{third}\mathrm{quadrant}. \\ \text{W}e\mathrm{know}\mathrm{that}\sin \text{e}\mathrm{function}\mathrm{is}\mathrm{positive}\mathrm{in}\text{the}\mathrm{second}\mathrm{quadrant}\mathrm{and} \\ \mathrm{in}\text{the}\mathrm{third}\mathrm{quadrant},\mathrm{both}\sin \text{e}\mathrm{and}\cos \text{ine}\mathrm{functions}\mathrm{are}\mathrm{negative}. \\ \mathrm{Therefore}, \\ \sin A=\sqrt{1-{\cos }^{2}A}=\sqrt{1-{\left(\frac{-12}{13}\right)}^2}=\sqrt{1-\frac{144}{169}}=\sqrt{\frac{25}{169}}=\frac{5}{13} \\ \sin B=-\frac{1}{\sqrt{1+{\cot }^{2}B}}=-\frac{1}{\sqrt{1+{\left({\displaystyle \frac{24}{7}}\right)}^2}}=\frac{-1}{\sqrt{1+{\displaystyle \frac{576}{49}}}}=\frac{-1}{\sqrt{{\displaystyle \frac{625}{49}}}}=\frac{-7}{25} \\ \cos B=-\sqrt{1-{\sin }^{2}B}=-\sqrt{1-{\left(\frac{-7}{25}\right)}^2}=-\sqrt{1-\frac{49}{625}}=-\sqrt{\frac{576}{625}}=-\frac{24}{25} \\ \mathrm{Now}, \\ \left(\mathrm{i}\right)\sin \left(A+B\right)=\sin A\cos B+\cos A+\sin B \\ =\frac{5}{13}\times \frac{-24}{25}+\frac{-12}{13}\times \frac{-7}{25} \\ =\frac{-120}{325}+\frac{84}{325} \\ =\frac{-36}{325} \\ \left(\mathrm{ii}\right)\cos \left(A+B\right)=\cos A\cos B-\sin A\sin B \\ =\frac{-12}{13}\times \frac{-24}{25}-\frac{5}{13}\times \frac{-7}{25} \\ =\frac{288}{325}+\frac{35}{325} \\ =\frac{323}{325} \\ \left(\mathrm{iii}\right)\tan \left(A+B\right)=\frac{\sin \left(A+B\right)}{\cos \left(A+B\right)}=\frac{{\displaystyle \raisebox{1ex}{$-36$}\!\left/ \!\raisebox{-1ex}{$325$}\right.}}{{\displaystyle \raisebox{1ex}{$323$}\!\left/ \!\raisebox{-1ex}{$325$}\right.}}=-\frac{36}{323} \end{gathered}$$