If `A={x in [0,2 pi]: tan x-tan^(2)x>0}` and `B={x in [0,2 pi]: |sin x|<(1)/(2)}`, then `A nn B=`

Solve sin x tan x -sin x+ tan x-1=0 for x in [0, 2pi] .

Solve tan x+tan 2x+tan 3x = tan x tan 2x tan 3x, x in [0, pi] .

Solve tan x>cot x, where x in[0,2 pi]

If x in [0,2pi] and |sec x+ tan x|=|sec x|+|tan x| , then x lies in the interval

If s={x in[0,2 pi]:det[[0,cos x,-sin xsin x,0,cos xcos x,sin x,0]]=}then sum_(x in s)tan((pi)/(3)+x) is equal to:

If Det[[1,sin x,sin^(2)x1,cos x,cos^(2)x1,tan x,tan^(2)x]]=0,x in[0,2 pi]

The number of solution of the equation tan3x-tan2x-tan3x.tan2x=1 in [0,2 pi] is

Solve tan x+tan2x+tan3x=tan x tan2x tan3x,x in[0,pi]

Prove that tan x gt x for all x in [0, (pi)/(2)]