Calculate tan(x-y), if sin x=1/2 and sin y=1/3. 0<x<90 90<y<180

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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 =...

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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)

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