If tan theta + tan 2theta + tan 3theta = tan | vedclass

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(B) Given the equation: $	an 	heta + 	an 2	heta + 	an 3	heta = 	an 	heta 	an 2	heta 	an 3	heta$. Rearranging the terms,we get: $	an 	het

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$\frac{n\pi}{6}$

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(B) Given the equation: $	an 	heta + 	an 2	heta + 	an 3	heta = 	an 	heta 	an 2	heta 	an 3	heta$. Rearranging the terms,we get: $	an 	heta + 	an 2	heta + 	an 3	heta - 	an 	heta 	an 2	heta 	an 3	heta = 0$. We know the identity: $	an(A+B+C) = \frac{	an A + 	an B + 	an C - 	an A 	an B 	an C}{1 - (	an A 	an B + 	an B 	an C + 	an C 	an A)}$. Let $A = 	heta, B = 2	heta, C = 3	heta$. Then $A+B+C = 6	heta$. The numerator of $	an(6	heta)$ is exactly the expression given in the problem. Since the expression equals $0$,we have $	an(6	heta) = 0$. The general solution for $	an x = 0$ is $x = n\pi$. Therefore,$6	heta = n\pi$,which implies $	heta = \frac{n\pi}{6}$.

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

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