We are given that sin x = (1/2) and sin y =(1/3). We have to find tan ( x - y).

sin x = (1/2)

=> tan x = (1/2)/ sqrt [1 - (1/2)^2]

=> (1/2)/ sqrt (3/4)

=> 1/ sqrt 3

sin y = (1/3)

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

=> (1/3)/sqrt (8/9)

=> 1/ sqrt 8

As 90 < y< 180

tan y = -1/ sqrt 8

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

=> [(1/ sqrt 3) + (1/ sqrt 8)] / [1 - (1/ sqrt 3)*(1/ sqrt 8)]

=> [(sqrt 8 + sqrt 3)/ sqrt 3*sqrt 8] / [(sqrt 3 * sqrt 8 - 1)/sqrt 3*sqrt 8

=> (sqrt 8 + sqrt 3)/ (sqrt 3*sqrt 8 -1 )

**tan (x - y) = (sqrt 8 + sqrt 3)/ (sqrt 3*sqrt 8 - 1)**

We'll write the formula of the tangent of difference of 2 angles.

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

Now, we'll have to establish the signature of tan x a and tan y. We know from enunciation that, tan x belongs to the first quadrant and it is has 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 - (1/2)^2]

cos x = sqrt(1 - 1/4)

cos x = sqrt3/2

tan x = (1/2)/(sqrt3/2)

tan x = sqrt 3/3

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

tan y = -(1/3)/sqrt(8/9)

tan y = -1/2sqrt2

tan y = -(sqrt2)/4

**tan (x-y) = [(sqrt 3)/3 + (sqrt2)/4]/[1 - (sqrt6)/12]**