Question

The period of $\left(\tan \theta -\frac{1}{3}{\tan }^3\theta \right){\left(\frac{1}{3}-{\tan }^2\theta \right)}^{-1}$ where ${\tan }^2\theta \ne \frac{1}{3}$ is

• A. $\frac{\pi }{3}$
• B. $\frac{2\pi }{3}$
• C. $\pi$
• D. $2\pi$

Answer 1

Answer: (A)

$\left(\tan \theta -\frac{1}{3}{\tan }^3\theta \right){\left(\frac{1}{3}-{\tan }^2\theta \right)}^{-1}$
where, ${\tan }^2\theta \ne \frac{1}{3}$
$=\frac{\left(\tan \theta -\frac{1}{3}{\tan }^3\theta \right)}{\left(\frac{1}{3}-{\tan }^2\theta \right)}=\frac{3\left(3\tan \theta -{\tan }^3\theta \right)}{3\left(1-3{\tan }^2\theta \right)}$
$=\tan 3\theta$
$\because \tan 3\theta =\left(\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\right)$
$=\tan (\pi +3\theta )$
$=\tan 3\left(\frac{\pi }{3}+\theta \right)$
So, the period of the given function is $\frac{\pi }{3}$.