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

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(C) Given equation is $	an 	heta + 	an 2	heta + \sqrt{3} 	an 	heta 	an 2	heta = \sqrt{3}.$Rearranging the terms,we get $	an 	heta + 	an 2

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$	heta = (3n + 1)\pi / 9, \forall n \in I$

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(C) Given equation is $	an 	heta + 	an 2	heta + \sqrt{3} 	an 	heta 	an 2	heta = \sqrt{3}.$Rearranging the terms,we get $	an 	heta + 	an 2	heta = \sqrt{3}(1 - 	an 	heta 	an 2	heta).$Dividing both sides by $(1 - 	an 	heta 	an 2	heta),$ we obtain $\frac{	an 	heta + 	an 2	heta}{1 - 	an 	heta 	an 2	heta} = \sqrt{3}.$Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B},$ we get $	an(3	heta) = \sqrt{3}.$Since $	an(\pi / 3) = \sqrt{3},$ we have $	an(3	heta) = 	an(\pi / 3).$The general solution is $3	heta = n\pi + \frac{\pi}{3}, \forall n \in I.$Dividing by $3,$ we get $	heta = \frac{n\pi}{3} + \frac{\pi}{9} = (3n + 1)\frac{\pi}{9}, \forall n \in I.$

If $\tan \theta + \tan 2\theta + \sqrt{3} \tan \theta \tan 2\theta = \sqrt{3},$ then

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