If tan theta + tan 2theta + sqrt{3} tan theta | vedclass

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(B) Given equation: $	an 	heta + 	an 2	heta + \sqrt{3} 	an 	heta 	an 2	heta = \sqrt{3}$ Rearranging the terms: $	an 	heta + 	an 2	heta = \

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$(3n+1) \frac{\pi}{9}, n \in Z$

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(B) Given equation: $	an 	heta + 	an 2	heta + \sqrt{3} 	an 	heta 	an 2	heta = \sqrt{3}$ Rearranging the terms: $	an 	heta + 	an 2	heta = \sqrt{3} - \sqrt{3} 	an 	heta 	an 2	heta$ $	an 	heta + 	an 2	heta = \sqrt{3}(1 - 	an 	heta 	an 2	heta)$ Dividing both sides by $(1 - 	an 	heta 	an 2	heta)$: $\frac{	an 	heta + 	an 2	heta}{1 - 	an 	heta 	an 2	heta} = \sqrt{3}$ Using the identity $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we get: $	an(3	heta) = \sqrt{3}$ Since $	an(\frac{\pi}{3}) = \sqrt{3}$,we have $	an(3	heta) = 	an(\frac{\pi}{3})$ The general solution is $3	heta = n\pi + \frac{\pi}{3}, n \in Z$ Dividing by $3$: $	heta = \frac{n\pi}{3} + \frac{\pi}{9} = (3n+1) \frac{\pi}{9}, n \in Z$ Thus,option $(B)$ is correct.

If $\tan \theta + \tan 2\theta + \sqrt{3} \tan \theta \tan 2\theta = \sqrt{3}$,then the general values of $\theta$ are

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