If cosx + cosy =1/8 and sinx + siny = 1/4 what is the value of tan(x+y) and tan[(x+y)/2]?

1) The following 3 results will be useful here in this question:

i) sin(C) + sin(D) = 2*sin{(C + D)/2}*cos{(C - D)/2} [Sum-Product rule]

ii) cos(C) + cos(D) = 2*cos{(C + D)/2}*cos{(C - D)/2} [Sum-Product rule]

iii) tan(2A) = 2tan(A)/(1 - tan²A) [Multiple angle identity]

2) So of the above, sin(x) + sin(y) = 2*sin{(x + y)/2}*cos{(x - y)/2} = 1/4

cos(x) + cos(y) = 2*cos{(x + y)/2}*cos{(x - y)/2} = 1/8

Dividing both the above, tan{(x + y)/2} = 2

3) Then applying the 3rd result, tan(x + y) = 2*tan{(x + y)/2}/[1 - tan²{(x + y)/2}]

Substituting the values of tan{(x + y)/2} = 2 from the above,

tan(x + y) = 2*2/(1 - 2²) = 4/(1 - 4) = -4/3

Thus, tan(x + y) = -4/3

tan{(x + y)/2} = 2