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

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(B) Given equation: $\sqrt{3} 	an 2	heta + \sqrt{3} 	an 3	heta + 	an 2	heta 	an 3	heta = 1$Rearranging the terms: $\sqrt{3}(	an 2	heta + 	a

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$\left(n + \frac{1}{6}\right)\frac{\pi}{5}$

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(B) Given equation: $\sqrt{3} 	an 2	heta + \sqrt{3} 	an 3	heta + 	an 2	heta 	an 3	heta = 1$Rearranging the terms: $\sqrt{3}(	an 2	heta + 	an 3	heta) = 1 - 	an 2	heta 	an 3	heta$Dividing both sides by $\sqrt{3}(1 - 	an 2	heta 	an 3	heta)$,we get:$\frac{	an 2	heta + 	an 3	heta}{1 - 	an 2	heta 	an 3	heta} = \frac{1}{\sqrt{3}}$Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have:$	an(2	heta + 3	heta) = 	an\left(\frac{\pi}{6}ight)$$	an 5	heta = 	an\left(\frac{\pi}{6}ight)$The general solution for $	an x = 	an \alpha$ is $x = n\pi + \alpha$.Therefore,$5	heta = n\pi + \frac{\pi}{6}$$	heta = \frac{n\pi}{5} + \frac{\pi}{30} = \left(n + \frac{1}{6}ight)\frac{\pi}{5}$.

If $\sqrt{3} \tan 2\theta + \sqrt{3} \tan 3\theta + \tan 2\theta \tan 3\theta = 1$,then the general value of $\theta$ is

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