The period of $\left(\tan \theta-\frac{1}{3} \tan ^3 \theta\right)\left(\frac{1}{3}-\tan ^2…
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
$\frac{\pi}{3}$
$\frac{2 \pi}{3}$
$\pi$
$2 \pi$
Solution
$\left(\tan \theta-\frac{1}{3} \tan ^3 \theta\right)\left(\frac{1}{3}-\tan ^2 \theta\right)^{-1}$
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 $\frac{\pi}{3}$.