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Trigonometrical Equations
easy
જો $\frac{{\tan 3\theta - 1}}{{\tan 3\theta + 1}} = \sqrt 3 $, તો $\theta $ નો વ્યાપક ઉકેલ મેળવો.
A
$\frac{{n\pi }}{3} + \frac{\pi }{{12}}$
B
$\frac{{n\pi }}{3} + \frac{{7\pi }}{{36}}$
C
$n\pi + \frac{{7\pi }}{{12}}$
D
$n\pi + \frac{\pi }{{12}}$
Solution
(b) $\frac{{\tan 3\theta – 1}}{{\tan 3\theta + 1}} = \sqrt 3 $
==> $\frac{{\tan 3\theta – \tan (\pi /4)}}{{1 + \tan 3\theta \,.\,\tan (\pi /4)}} = \sqrt 3 $
==> $\tan \,\left( {3\theta – \frac{\pi }{4}} \right) = \tan \,\frac{\pi }{3}$
==> $3\theta – (\pi /4) = n\pi + (\pi /3)$
==> $3\theta = n\pi + \frac{{7\pi }}{{12}}$
==> $\theta = \frac{{n\pi }}{3} + \frac{{7\pi }}{{36}}$.
Standard 11
Mathematics
