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The value of $\frac{\tan x}{\tan 3 x}$ whenever defined never lie between
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Verified Answer
The correct answer is:
$1 / 3$ and 3
$\begin{aligned}
& y=\frac{\tan x}{\tan 3 x}=\frac{\tan x}{3 \tan x-\tan ^3 x} \\
& \text { Let } \quad 1-3 \tan ^2 x \\
& y=\frac{1-3 \tan ^2 x}{3-\tan ^2 x}=\frac{\frac{1}{3}-\tan ^2 x}{1-\frac{1}{3} \cdot \tan ^2 x} \\
&
\end{aligned}$
Hence, $y$ should never lie between $\frac{1}{3}$ and 3 whenever defined.
& y=\frac{\tan x}{\tan 3 x}=\frac{\tan x}{3 \tan x-\tan ^3 x} \\
& \text { Let } \quad 1-3 \tan ^2 x \\
& y=\frac{1-3 \tan ^2 x}{3-\tan ^2 x}=\frac{\frac{1}{3}-\tan ^2 x}{1-\frac{1}{3} \cdot \tan ^2 x} \\
&
\end{aligned}$
Hence, $y$ should never lie between $\frac{1}{3}$ and 3 whenever defined.
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