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  4. The period of ( tan θ-(1/3) tan 3 θ)((1/3)- tan 2 θ)-1 where tan 2 θ ≠ (1/3) is

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Q. 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

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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 $\frac{\pi}{3}$.