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The period of
$\left(\tan \theta-\frac{1}{3} \tan ^3 \theta\right)\left(\frac{1}{3}-\tan ^2 \theta\right)^{-1}$
where $\tan ^2 \theta \neq \frac{1}{3}$ is
Options:
$\left(\tan \theta-\frac{1}{3} \tan ^3 \theta\right)\left(\frac{1}{3}-\tan ^2 \theta\right)^{-1}$
where $\tan ^2 \theta \neq \frac{1}{3}$ is
Solution:
1658 Upvotes
Verified Answer
The correct answer is:
$\frac{\pi}{3}$
$\left(\tan \theta-\frac{1}{3} \tan ^3 \theta\right)\left(\frac{1}{3}-\tan ^2 \theta\right)^{-1}$
where, $\tan ^2 \theta \neq \frac{1}{3}$
$=\frac{\left(\tan \theta-\frac{1}{3} \tan ^3 \theta\right)}{\left(\frac{1}{3}-\tan ^2 \theta\right)}=\frac{3\left(3 \tan \theta-\tan ^3 \theta\right)}{3\left(1-3 \tan ^2 \theta\right)}$
$=\tan 3 \theta$
$\because \tan 3 \theta=\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right)$
$=\tan (\pi+3 \theta)$
$=\tan 3\left(\frac{\pi}{3}+\theta\right)$
So, the period of the given function is $\frac{\pi}{3}$.
where, $\tan ^2 \theta \neq \frac{1}{3}$
$=\frac{\left(\tan \theta-\frac{1}{3} \tan ^3 \theta\right)}{\left(\frac{1}{3}-\tan ^2 \theta\right)}=\frac{3\left(3 \tan \theta-\tan ^3 \theta\right)}{3\left(1-3 \tan ^2 \theta\right)}$
$=\tan 3 \theta$
$\because \tan 3 \theta=\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right)$
$=\tan (\pi+3 \theta)$
$=\tan 3\left(\frac{\pi}{3}+\theta\right)$
So, the period of the given function is $\frac{\pi}{3}$.
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