Question 927606
If we need exact values,  we need to use the trigonometric identities
{{{cos(A+B)=cos(A)*cos(B)-sin(A)*sin(B)}}} and {{{sin(A+B)=cos(A)*sin(B)+sin(A)*cos(B)}}} ,
so as a first step, we are going to find exact values for {{{sin(A)}}} and {{{cos(b)}}} .
 
{{{cos(A)=4/9}}} in Q1.
In quadrant 1, sine and cosine are positive , and we know that {{{(cos(A))^2+(sin(A))^2=1}}} , so
{{{sin(A)=sqrt(1-(4/9)^2)=sqrt(1-16/81)=sqrt((81-16)/81)=sqrt(65/81)=sqrt(65)/9}}}
 
{{{tan(B)=-5/4}}} in Q2 means
{{{drawing(300,300,-7,3,-3,7,grid(0),
red(triangle(0,0,-4,5,-4,0)),
red(rectangle(-4,0,-3.8,0.2)),
red(arrow(0,0,-4.8,6)),
red(arc(0,0,4.05,4.05,-128.66,0)),
red(triangle(-1.249,1.5617,-1.249,1.76,-1.05,1.5617)),
locate(0.5,1.8,red(B)),locate(-3.9,2.6,red(5)),
locate(-2.2,0.6,red(4)),locate(-2.7,2.7,red(sqrt(41)))
)}}} From the red right rectangle, we calculate
{{{sin(B)=5/sqrt(41)=5sqrt(41)/41}}} and {{{cos(B)=-4/sqrt(41)=-4sqrt(41)/41}}} .
 
Now,
{{{cos(A+B)=(4/9)*(-4sqrt(41)/41)-(sqrt(65)/9)*(5sqrt(41)/41) =(-16sqrt(41)-5sqrt(41)*sqrt(65))/(9*41)=highlight(-(16sqrt(41)+5sqrt(2665))/369)}}}
{{{sin(A+B)=(4/9)*(5sqrt(41)/41)+(sqrt(65)/9)*(-4sqrt(41)/41)=(20sqrt(41)-4sqrt(41)*sqrt(65))/(9*41)=(20sqrt(41)-4sqrt(2665))/369}}}
{{{tan(A+B)=sin(A+B)/cos(A+B)}}}={{{highlight((20sqrt(41)-4sqrt(41*65))/(-(16sqrt(41)+5sqrt(41*65))))}}}={{{-4(5sqrt(41)-sqrt(41*65))(16sqrt(41)-5sqrt(41*65))/((16sqrt(41)+5sqrt(41*65))(16sqrt(41)-5sqrt(41*65)))}}}={{{-4(80*41-25*41sqrt(65)-16*41sqrt(65)+5*41*65)/(256*41-25*41*65)}}}
 
={{{-4*41*(80-25sqrt(65)-16sqrt(65)+5*65)/(41*(256-25*65))}}}={{{-4*(80-25sqrt(65)-16sqrt(65)+5*65)/(256-1625)}}}={{{-4*(80-41sqrt(65)+325)/(256-1625)}}}
 
={{{-4*(405-41sqrt(65))/(-1369)}}}={{{highlight(1620-164sqrt(65))/1369)}}}