Given sin x = 2/3 and sin y=1/3, 0

First of all, we'll have to find out if tan x and tan y
are positive or negative. From enunciation, we'll have tan x belongs to the first
quadrant and it is positive and tan y belongs to the second quadrant and it is
negative.

tan x=sin x/cos
x

cos x = sqrt[1 - (sin
x)^2]

cos x = sqrt[1 -
(2/3)^2]

cos x = sqrt (1 -
4/9)

cos x = (sqrt 5)/3

cos y
= - sqrt[1 - (sin y)^2]

cos y = - sqrt[1 -
(1/3)^2]

cos y = - sqrt[1 -
(1/9)]

cos y = - 2(sqrt
2)/3

tan x= (2/3)/[(sqrt
5)/3]

tan x = 2(sqrt5)/5

tan
y= (1/3)/[- 2(sqrt 2)/3]

tan y =
(-sqrt2)/4

tan (x+y)=(tan x + tan y)/(1-tan x*tan
y)

tan (x+y)=[2(sqrt5)/5+
(-sqrt2)/4]/[1+2(sqrt10)/20]

tg (x+y)=
[8(sqrt5) - 5(sqrt2)]/[20+2(sqrt10)]