Solution
Given that
[tex]\begin{gathered} \cos A=\frac{8}{17}\text{ and } \\ \sin B=\frac{5}{13} \end{gathered}[/tex]
Let's draw the two diagrams for angles A and B
Hence,
[tex]\begin{gathered} \sin A=\frac{15}{17} \\ \cos B=\frac{12}{13} \\ \tan A=\frac{15}{8} \\ \tan B=\frac{5}{12} \end{gathered}[/tex]
Therefore,
[tex]\sin (A+B)=\sin A\cos B+\sin B\cos A=\frac{15}{17}\times\frac{12}{13}+\frac{5}{13}\times\frac{8}{17}=\frac{220}{221}[/tex][tex]\sin (A-B)=\sin A\cos B-\sin B\cos A=\frac{15}{17}\times\frac{12}{13}-\frac{5}{13}\times\frac{8}{17}=\frac{140}{221}[/tex][tex]\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{\frac{15}{8}+\frac{5}{12}}{1-\frac{15}{8}\times\frac{5}{12}}=\frac{220}{21}[/tex][tex]\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}=\frac{\frac{15}{8}-\frac{5}{12}}{1+\frac{15}{8}\times\frac{5}{12}}=\frac{140}{171}[/tex]
