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The set of values of x for which tan3x-tan2x/1+tan3xtan2x=1 is
WE KNOW
tanx = tan(3x-2x) = (tan3x - tan2x)/(1+tan3xtan2x)
therefore the above problem reduce to
tanx=1
then find x.
(tan 3x - tan 2x) / (1 + tan 3x tan 2x) = tan (3x-2x) = tan x
So the equation becomes tan x = 1 = tan (pi/4)
Whose general solution is x = n*pi + pi/4 = (n+1/4) * pi
tan(3x-2x)=1
x=n*pi + (pi/4)
hi pavan
the Kaushik is correct
you just have to apply tan ( A - B ) formula
tan (A-B) = (tanA - tan B )/(1 - tanA tan B )
take A = 3 x and B = 2x
jitender
askiitians expert
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