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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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$\large\frac{\tan 3x}{\tan x}$ lies between

$\begin{array}{1 1}(a)\;-\large\frac{1}{3}\normalsize\;and\;\;0\\(b)\;\;\large\frac{1}{3}\normalsize\;and\;\;3\\(c)\;-\large\frac{1}{3}\normalsize\;and\;\;\large\frac{1}{3}\\(d)\;-3\;and\;\;3\end{array}$

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1 Answer

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Let $y=\large\frac{\tan 3x}{\tan x}$
Then,$y=\large\frac{3\tan x-\tan^3x}{\tan x(1-3\tan^2x)}$
$y=\large\frac{\tan x(3-\tan^2x)}{\tan x(1-3\tan^2x)}$
$y=\large\frac{(3-\tan^2x)}{(1-3\tan^2x)}$
$\tan^2x(3y-1)=y-3$
$\tan^2x=\large\frac{y-3}{3y-1}$
Now $\tan^2x\geq 0$ for all $x$
$\large\frac{y-3}{3y-1}$$\geq 0$
$(y-3)(3y-1)\geq 0$
$y\leq \large\frac{1}{3}$ (or) $y\geq 3$
$\Rightarrow y$ does not lie between $\large\frac{1}{3}$ and $3$
Hence (b) is the correct option.
answered Oct 14, 2013 by sreemathi.v
 

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