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 = (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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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]
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