The period of left(tan theta - frac{1}{3} tan | vedclass

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Question

(A) Let $f(	heta) = \left(	an 	heta - \frac{1}{3} 	an^3 	hetaight) \left(\frac{1}{3} - 	an^2 	hetaight)^{-1}$.Given $	an^2 	heta 
eq \fr

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(A) Let $f(	heta) = \left(	an 	heta - \frac{1}{3} 	an^3 	hetaight) \left(\frac{1}{3} - 	an^2 	hetaight)^{-1}$.Given $	an^2 	heta 
eq \frac{1}{3}$,we can rewrite the expression as:$f(	heta) = \frac{	an 	heta - \frac{1}{3} 	an^3 	heta}{\frac{1}{3} - 	an^2 	heta} = \frac{\frac{3 	an 	heta - 	an^3 	heta}{3}}{\frac{1 - 3 	an^2 	heta}{3}} = \frac{3 	an 	heta - 	an^3 	heta}{1 - 3 	an^2 	heta}$.Using the identity $	an(3	heta) = \frac{3 	an 	heta - 	an^3 	heta}{1 - 3 	an^2 	heta}$,we get $f(	heta) = 	an(3	heta)$.The period of $	an(x)$ is $\pi$. Therefore,the period of $	an(3	heta)$ is $\frac{\pi}{3}$.

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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