Question

The value of $\tan \mathrm{\theta }+\tan \left(60°+\mathrm{\theta }\right)+\tan \left(120°+\mathrm{\theta }\right)$ is
(a) $3\tan 3\mathrm{\theta }$
(b) $\tan 3\mathrm{\theta }$
(c) $3\cot 3\mathrm{\theta }$
(d) $\cot 3\mathrm{\theta }$

Answer 1

(a) $3\tan 3\mathrm{\theta }$

$$\begin{gathered} \tan \mathrm{\theta }+\tan \left(60°+\mathrm{\theta }\right)+\tan \left(120°+\mathrm{\theta }\right)=\tan \mathrm{\theta }+\frac{\tan 60°+\mathrm{tan\theta }}{1-\tan 60°\mathrm{tan\theta }}+\frac{\tan 120°+\tan \mathrm{\theta }}{1-\tan 120°\mathrm{tan\theta }} \\ =\tan \theta +\frac{\sqrt{3}+\tan \mathrm{\theta }}{1-\sqrt{3}\mathrm{tan\theta }}+\frac{\left(-\sqrt{3}+\mathrm{tan\theta }\right)}{1+\sqrt{3}\mathrm{tan\theta }} \\ =\frac{\tan \theta \left(1-3{\tan }^2\mathrm{\theta }\right)+\left(\sqrt{3}+\mathrm{tan\theta }\right)\left(1+\sqrt{3}\mathrm{tan\theta }\right)+\left(-\sqrt{3}+\mathrm{tan\theta }\right)\left(1-\sqrt{3}\mathrm{tan\theta }\right)}{1-3{\tan }^2\mathrm{\theta }} \\ =\frac{\tan \theta -3{\tan }^3\mathrm{\theta }+\sqrt{3}+3\tan \theta +\mathrm{tan\theta }+\sqrt{3}{\tan }^2\mathrm{\theta }+\mathrm{tan\theta }-\sqrt{3}{\tan }^2\mathrm{\theta }-\sqrt{3}+3\mathrm{tan\theta }}{1-3{\tan }^2\mathrm{\theta }} \\ =\frac{9\mathrm{tan\theta }-3{\tan }^3\mathrm{\theta }}{1-3{\tan }^2\mathrm{\theta }} \\ =\frac{3\left(3\mathrm{tan\theta }-{\tan }^3\mathrm{\theta }\right)}{1-3{\tan }^2\mathrm{\theta }} \\ =3\tan 3\mathrm{\theta } \end{gathered}$$