If tan theta + tan left(theta + frac{pi}{3}ri | vedclass

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(B) Given,$	an 	heta + 	an \left(	heta + \frac{\pi}{3}ight) + 	an \left(	heta + \frac{2\pi}{3}ight) = 3$. Using the identity $	an(A+B) = \f

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$	an 3	heta$

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(B) Given,$	an 	heta + 	an \left(	heta + \frac{\pi}{3}ight) + 	an \left(	heta + \frac{2\pi}{3}ight) = 3$. Using the identity $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have: $	an 	heta + \frac{	an 	heta + \sqrt{3}}{1 - \sqrt{3} 	an 	heta} + \frac{	an 	heta - \sqrt{3}}{1 + \sqrt{3} 	an 	heta} = 3$. Combining the fractions: $	an 	heta + \frac{(	an 	heta + \sqrt{3})(1 + \sqrt{3} 	an 	heta) + (	an 	heta - \sqrt{3})(1 - \sqrt{3} 	an 	heta)}{1 - 3 	an^2 	heta} = 3$. $	an 	heta + \frac{(	an 	heta + \sqrt{3} 	an^2 	heta + \sqrt{3} + 3 	an 	heta) + (	an 	heta - \sqrt{3} 	an^2 	heta - \sqrt{3} + 3 	an 	heta)}{1 - 3 	an^2 	heta} = 3$. $	an 	heta + \frac{8 	an 	heta}{1 - 3 	an^2 	heta} = 3$. $\frac{	an 	heta - 3 	an^3 	heta + 8 	an 	heta}{1 - 3 	an^2 	heta} = 3$. $\frac{9 	an 	heta - 3 	an^3 	heta}{1 - 3 	an^2 	heta} = 3$. $3 \left( \frac{3 	an 	heta - 	an^3 	heta}{1 - 3 	an^2 	heta} ight) = 3$. Since $	an 3	heta = \frac{3 	an 	heta - 	an^3 	heta}{1 - 3 	an^2 	heta}$,we get $3 	an 3	heta = 3$,which implies $	an 3	heta = 1$. Thus,option $B$ is correct.

If $\tan \theta + \tan \left(\theta + \frac{\pi}{3}\right) + \tan \left(\theta + \frac{2\pi}{3}\right) = 3$,then which of the following is equal to $1$?

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