Given:-
[tex]\tan A=\frac{21}{20},\cos B=\frac{15}{17}[/tex]To find:-
[tex]\sin (A+B)[/tex]The formula is,
[tex]\sin (A+B)=\sin A\cos B+\cos A\sin B[/tex]Now we substitute the values. we get,
[tex]\sin (A+B)=21\times\frac{15}{17}+20\sqrt[]{(1-\frac{15^2}{17^{^2^{}}})}[/tex]Now we get,
[tex]\begin{gathered} \sin (A+B)=21\times\frac{15}{17}+20\sqrt[]{(1-\frac{15^2}{17^{^2^{}}})} \\ \sin (A+B)=\frac{315}{17}+\frac{20}{17}\sqrt[]{17^2-15^2} \\ \sin (A+B)=\frac{315}{17}+\frac{20}{17}\times8 \\ \sin (A+B)=\frac{315}{17}+\frac{160}{17} \\ \sin (A+B)=\frac{475}{17} \end{gathered}[/tex]So the required value is,
[tex]\frac{475}{17}[/tex]