Question

The value of $\tan \mathrm{\theta }\tan \left(60°-\mathrm{\theta }\right)\tan \left(60°+\mathrm{\theta }\right)$ is
(a) $\cot 3\mathrm{\theta }$
(b) $2\cot 3\mathrm{\theta }$
(c) $\tan 3\mathrm{\theta }$
(d) $3\tan 3\mathrm{\theta }$

Answer 1

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

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