Use the given information to find (a) sin(s+t), (b) tan(s+t), and (c) the quadrant of s+t. cos s = - 12/13 and sin t = 4/5, s and t in quadrant ll.

Answer:Part a) [tex]sin(s + t) =-\frac{63}{65}[/tex]    Part b) [tex]tan(s + t) = -\frac{63}{16}[/tex]Part c) (s+t) lie on Quadrant IVStep-by-step explanation:[Part a) Find sin(s+t)we know that[tex]sin(s + t) = sin(s) cos(t) + sin(t)cos(s)[/tex]step 1Find sin(s)[tex]sin^{2}(s)+cos^{2}(s)=1[/tex]we have[tex]cos(s)=-\frac{12}{13}[/tex]substitute[tex]sin^{2}(s)+(-\frac{12}{13})^{2}=1[/tex][tex]sin^{2}(s)+(\frac{144}{169})=1[/tex][tex]sin^{2}(s)=1-(\frac{144}{169})[/tex][tex]sin^{2}(s)=(\frac{25}{169})[/tex][tex]sin(s)=\frac{5}{13}[/tex] ---> is positive because s lie on II Quadrantstep 2Find cos(t)[tex]sin^{2}(t)+cos^{2}(t)=1[/tex]we have[tex]sin(t)=\frac{4}{5}[/tex]substitute[tex](\frac{4}{5})^{2}+cos^{2}(t)=1[/tex][tex](\frac{16}{25})+cos^{2}(t)=1[/tex][tex]cos^{2}(t)=1-(\frac{16}{25})[/tex][tex]cos^{2}(t)=\frac{9}{25}[/tex][tex]cos(t)=-\frac{3}{5}[/tex] is negative because t lie on II Quadrantstep 3Find sin(s+t)[tex]sin(s + t) = sin(s) cos(t) + sin(t)cos(s)[/tex]we have[tex]sin(s)=\frac{5}{13}[/tex] [tex]cos(t)=-\frac{3}{5}[/tex][tex]sin(t)=\frac{4}{5}[/tex][tex]cos(s)=-\frac{12}{13}[/tex]substitute the values[tex]sin(s + t) = (\frac{5}{13})(-\frac{3}{5}) + (\frac{4}{5})(-\frac{12}{13})[/tex][tex]sin(s + t) = -(\frac{15}{65}) -(\frac{48}{65})[/tex][tex]sin(s + t) =-\frac{63}{65}[/tex]Part b) Find tan(s+t)we know thattex]tan(s + t) = (tan(s) + tan(t))/(1 - tan(s)tan(t))[/tex]we have[tex]sin(s)=\frac{5}{13}[/tex] [tex]cos(t)=-\frac{3}{5}[/tex][tex]sin(t)=\frac{4}{5}[/tex][tex]cos(s)=-\frac{12}{13}[/tex]step 1Find tan(s)[tex]tan(s)=sin(s)/cos(s)[/tex]substitute[tex]tan(s)=(\frac{5}{13})/(-\frac{12}{13})=-\frac{5}{12}[/tex]step 2Find tan(t)[tex]tan(t)=sin(t)/cos(t)[/tex]substitute[tex]tan(t)=(\frac{4}{5})/(-\frac{3}{5})=-\frac{4}{3}[/tex]step 3Find tan(s+t)[tex]tan(s + t) = (tan(s) + tan(t))/(1 - tan(s)tan(t))[/tex]substitute the values[tex]tan(s + t) = (-\frac{5}{12} -\frac{4}{3})/(1 - (-\frac{5}{12})(-\frac{4}{3}))[/tex][tex]tan(s + t) = (-\frac{21}{12})/(1 - \frac{20}{36})[/tex][tex]tan(s + t) = (-\frac{21}{12})/(\frac{16}{36})[/tex][tex]tan(s + t) = -\frac{63}{16}[/tex]Part c) Quadrant of s+twe know that[tex]sin(s + t) =negative[/tex]  ----> (s+t) could be in III or IV quadrant[tex]tan(s + t) =negative[/tex] ----> (s+t) could be in III or IV quadrantFind the value of cos(s+t)[tex]cos(s+t) = cos(s) cos(t) -sin (s) sin(t)[/tex]we have[tex]sin(s)=\frac{5}{13}[/tex] [tex]cos(t)=-\frac{3}{5}[/tex][tex]sin(t)=\frac{4}{5}[/tex][tex]cos(s)=-\frac{12}{13}[/tex]substitute[tex]cos(s+t) = (-\frac{12}{13})(-\frac{3}{5})-(\frac{5}{13})(\frac{4}{5})[/tex][tex]cos(s+t) = (\frac{36}{65})-(\frac{20}{65})[/tex][tex]cos(s+t) =\frac{16}{65}[/tex]we have that[tex]cos(s+t)=positive[/tex] -----> (s+t) could be in I or IV quadrant[tex]sin(s + t) =negative[/tex]  ----> (s+t) could be in III or IV quadrant[tex]tan(s + t) =negative[/tex] ----> (s+t) could be in III or IV quadranttherefore(s+t) lie on Quadrant IV