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

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(b) $	an 	heta + 	an 2	heta + 	an 3	heta = 	an 	heta 	an 2	heta 	an 3	heta $

$	an 6	heta = \frac{{	an 	heta + 	an 2	heta + 	an 3

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) $	an 	heta + 	an 2	heta + 	an 3	heta = 	an 	heta 	an 2	heta 	an 3	heta $

$	an 6	heta = \frac{{	an 	heta + 	an 2	heta + 	an 3	heta - 	an 	heta 	an 2	heta 	an 3	heta }}{{1 - \sum 	an 	heta 	an 2	heta }}$

$= 0, $ (from the given condition)

$ \Rightarrow $ $6	heta = n\pi \Rightarrow 	heta = n\pi /6$.

Trick : In such type of problems, the general value of $	heta $ is given by $\frac{{n\pi }}{{{m{sum \,of\, number\, of\, }}	heta }}$.

So the general value of $	heta $ is $\frac{{n\pi }}{{1 + 2 + 3}} = \frac{{n\pi }}{6}$.

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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