If frac{{tan 3theta - 1}}{{tan 3theta + 1}} = | vedclass

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Question

(b) $\frac{{	an 3	heta - 1}}{{	an 3	heta + 1}} = \sqrt 3 $

==> $\frac{{	an 3	heta - 	an (\pi /4)}}{{1 + 	an 3	heta \,.\,	an (\pi /4)}}

Vedclass · Accepted Answer

$\frac{{n\pi }}{3} + \frac{{7\pi }}{{36}}$

Vedclass · Answer

(b) $\frac{{	an 3	heta - 1}}{{	an 3	heta + 1}} = \sqrt 3 $

==> $\frac{{	an 3	heta - 	an (\pi /4)}}{{1 + 	an 3	heta \,.\,	an (\pi /4)}} = \sqrt 3 $

==> $	an \,\left( {3	heta - \frac{\pi }{4}} ight) = 	an \,\frac{\pi }{3}$

==> $3	heta - (\pi /4) = n\pi + (\pi /3)$

==> $3	heta = n\pi + \frac{{7\pi }}{{12}}$

==> $	heta = \frac{{n\pi }}{3} + \frac{{7\pi }}{{36}}$.

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

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