question_answer

\[\frac{\tan 3x}{\tan x}\]ever lies between -

A) 2 and \[\frac{1}{2}\]                 

B) 3 and \[\frac{1}{3}\]

C) 4 and \[\frac{1}{4}\]                 

D) 5 and \[\frac{1}{5}\]

Correct Answer: B

Solution :

\[y=\frac{3\tan x-{{\tan }^{3}}x}{(1-3{{\tan }^{2}}x)\,\,\tan x}\]

\[y=\frac{3-{{\tan }^{2}}x}{1-3{{\tan }^{2}}x}\]

\[{{\tan }^{2}}x=\frac{3-y}{1-3y}\]

\[if\,\,{{\tan }^{2}}x\ge 0\]

\[\frac{3-y}{1-3y}\ge 0\]

Never lies between \[\frac{1}{3}\]and 3.