# Given sin x = 2/3 and sin y=1/3, 0<x<90, 90<y<180, find tan(x+y)?

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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)]**