Question

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

• A. $ \frac{n\pi}{3} -\frac{\pi}{12} $
• B. $ n\pi + \frac{7\pi}{12} $
• C. $ \frac{n\pi}{3} + \frac{7\pi}{36} $
• D. $ n\pi + \frac{\pi}{12} $

Answer 1

Answer: (C)

Given that, $\frac{tan 3\theta -1}{tan\, 3\theta +1} = \sqrt{3} $
$ \Rightarrow tan \,3\theta - 1 = \sqrt{3} \,tan\, 3\theta + \sqrt{3} $
$ \Rightarrow tan\, 3\theta -1 - \sqrt{3} tan\, 3\theta - \sqrt{3} = 0 $
$ \Rightarrow tan \,3\theta\left(1-\sqrt{3}\right) - \left(1+\sqrt{3}\right) = 0 $
$\Rightarrow tan \,3 \theta = \frac{1 +\sqrt{3}}{1-\sqrt{3}}$
$ \Rightarrow tan\, 3\theta = tan \,105^{\circ} = tan \frac{7\pi}{12} $
$ \therefore 3\theta = n \pi + \frac{7\pi}{12} $
$\Rightarrow \theta = \frac{n\pi}{3} + \frac{7\pi}{12} $