prove that the value of (sin x*cos 3x)/(cos x*sin 3x) cannot lie between 1/3 and 3.

it is tan x/ tan 3x

= tan x/ (tan x [ 3 - tan^2(x) ]/ [1 - 3tan^2(x) ])
= (1 -3 tan ^2 x)/(3 - tan^2 x )

say tan x = y which can take any value and

t = (1-3y^2)/(3-y^2)

or 3t - ty^2 = 1- 3y^2

or (3-t)y^2 + 3t - 1 = 0

or (3-t)^2y^2 = (3t-1)(t-3)

as LHS >=0 so RHS >=0

(3t-1)(t-3) >= 0 => t >= 1/3 and t >=3 => t >= 3

or t <= 1/3 and t <=3 => t <= 1/3

so cannot lie between 1/3 and 3.