mohon bantu ya materi trigonometri

a. diketahui
sinx = 3/5   ==>>   tanx = 3/4
cosy = -5/13   ====>>  tany = -12/5
ditanya : tan(x+y) dan tan(x-y)
(i)
tan(x+y) 
= (tanx + tany) / (1 - tanx . tany)
= (3/4 - 12/5) / ( 1 - (3/4)(-12/5))
= (-33/20) / (14/5)
= -33/56

(ii)
tan(x - y)
= (tanx - tany) / (1 + tanx . tany)
= (3/4 + 12/5) / ( 1 + (3/4)(-12/5))
= (63/20) / (-4/5)
= -63/16

jawab
(3)
tan (A+B) = 2
tan A = 1

tan (A+B) = (tan A + tan B) / (1 - tan A tan B)
2 = (1 + tan B) / (1 - (1)(tan B))
2 ( 1-  tan B) = 1 + tan B
2 - 2 tan B = 1 + tan B
- 2 tan B - tan B =1 - 2
- 3 (tan B) = - 1
tan  B = - 1/-3  = 1/3