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

tanx=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

We know that tan (A - B) = (tan A- tan B) / (1+ tan A. tanB) So (tan3x - tan2x) / (1+ tan3x tan2x) = tan (3x - 2x) = tan x So we have to find solution of tan x = 1 So x = p / 4 will 1 solution and x = np + p/4 will be general solution of equation. Thanks & Regards, Amit Nardiya, askIITians faculty.

We know that tan (A - B) = (tan A- tan B) / (1+ tan A. tanB) So (tan3x - tan2x) / (1+ tan3x tan2x) = tan (3x - 2x) = tan x So we have to find solution of tan x = 1 So x = p / 4 will 1 solution and x = np + p/4 will be general solution of equation.

The answer for the given question is phi (?) or a null set.. tan (A - B) = (tan A- tan B) / (1+ tan A. tanB) So (tan3x - tan2x) / (1+ tan3x tan2x) = tan (3x - 2x) = tan x So we have to find solution of tan x = 1 So x = p / 4 will 1 solution and x = np + p/4 will be general solution of equation. But if x= p / 4 then, 2x= p / 2. And tan2x will not be defined.. You can check for other values of x as well...So The answer is phi (?) or a null set.

The answer to this question is there is no set of solutions for this question, because if we take x=π/4 then if we substitute it in the equation we get an indefinite form of (infinity/infinity).... Hence there is no solution to this answer