Question

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

• A. $\frac{n\pi }{3}-\frac{\pi }{2}$
• B. $n\pi +\frac{7\pi }{2}$
• C. $\frac{n\pi }{3}-\frac{7\pi }{36}$
• D. $n\pi +\frac{\pi }{12}$

Answer 1

The correct option is C $\frac{n\pi }{3}-\frac{7\pi }{36}$

Given that, $\frac{\tan 3\theta -1}{\tan 3\theta +1}=\sqrt{3}$
$\Rightarrow \tan 3\theta -1=\sqrt{3}\tan 3\theta +\sqrt{3}$
$\Rightarrow \tan 3\theta -1-\sqrt{3}\tan 3\theta -\sqrt{3}=0$
$\Rightarrow \tan 3\theta (1-\sqrt{3})-(1+\sqrt{3})=0$
$\Rightarrow \tan 3\theta =\frac{1+\sqrt{3}}{1-\sqrt{3}}$
$=\frac{\tan {45}^{\circ }+\tan {60}^{\circ }}{1-\tan {45}^{\circ }.\tan {60}^{\circ }}$
$=\tan ({45}^{\circ }+{60}^{\circ })$
$\Rightarrow \tan 3\theta =\tan {105}^{\circ }=\tan \frac{7\pi }{12}$
$\therefore 3\theta =n\pi +\frac{7\pi }{12}\Rightarrow \theta =\frac{n\pi }{3}+\frac{7\pi }{36}$