`tan^(3)x-3tanx=0`

If tan of the angles A , B , C are the solutions of the equations tan^(3)x-3ktan^(2)x-3tanx+k=0 , then the triangle ABC is

If the angles A , B , C are the solutions of the equations tan^(3)-3ktan^(2)x-3tanx+k=0 , then the triangle ABC is

Prove that : tan 3x=(3tanx-tan^(3)x)/(1-3tan^(2)x)

If tan of angles A,B,C are the solutions of the equations tan^3x+3ktan^2x-3tanx+k=0 , then the triangle ABC is

lim_(x rarr tan^(-1)) (tan^(2)x - 2tanx -3)/(tan^(2)x - 4tan x +3 ) =

Evaluate: lim_(xto(pi)/(3))(tan^3x-3tanx)/(cos(x+(pi)/(6)))

lim_(x rarr 0) (tan^(-1)3x-4tanx)/(4 sin^(-1) 2x -7x) =

If tanx+tan(x+pi/3)+tan(x+(2pi)/3)=3 then prove that (3tanx-tan^3x)/(1-3tan^2x)=1

Prove that tan 3x=(3tanx-tan^3x)/(1-3tan^2x)

(d)/(dx)[tan^(-1)((2+3tanx)/(3-2tanx))]=