Question

If $tan\theta +tan\theta ({60}^0+\theta )+tan({120}^0+\theta )=3$ then $\theta$ =

• A. $\frac{n\pi }{3}+\frac{\pi }{12}$
• B. $n\pi +\frac{\pi }{4}$
• C. Both A and B
• D. None of these.

Answer 1

The correct option is A $\frac{n\pi }{3}+\frac{\pi }{12}$

Given,

$\tan \theta +\tan (60+\theta )+\tan (120+\theta )$

$=\tan \theta +\frac{\tan 60+\tan \theta }{1-\tan 60\tan \theta }+\frac{\tan 120+\tan \theta }{1-\tan 120\tan \theta }$

$=\tan \theta +\frac{\sqrt{3}+\tan \theta }{1-\sqrt{3}\tan \theta }+\frac{-\sqrt{3}+\tan \theta }{1+\sqrt{3}\tan \theta }$

upon solving the above equation, we get,

$=\tan \theta +\frac{8\tan \theta }{1-3{\tan }^2\theta }$

$=\frac{9\tan \theta -3{\tan }^3\theta }{1-3{\tan }^2\theta }$

$=3\tan 3\theta$

But, from given, we have,

$\tan \theta +\tan (60+\theta )+\tan (120+\theta )=3$

$\Rightarrow 3=3\tan 3\theta$

$3\theta =\frac{\pi }{4}+\pi n$

$\therefore \theta =\frac{\pi }{12}+\frac{n\pi }{3}$