Question

$\tan {20}^{\circ }\tan {40}^{\circ }\tan {80}^{\circ }=\tan {60}^{\circ }$

Answer 1

As we know that

$\sin \theta \sin ({60}^{\circ }-\theta )\sin ({60}^{\circ }+\theta )=\frac{1}{4}\sin 3\theta$

$\cos \theta \cos ({60}^{\circ }-\theta )\cos ({60}^{\circ }+\theta )=\frac{1}{4}\cos 3\theta$

$\tan \theta \tan ({60}^{\circ }-\theta )\tan ({60}^{\circ }+\theta )=\frac{\sin \theta }{\cos \theta }\times \frac{\sin ({60}^{\circ }-\theta )}{\cos ({60}^{\circ }-\theta )}\times \frac{\sin ({60}^{\circ }+\theta )}{\cos ({60}^{\circ }+\theta )}$

$=\frac{\sin \theta \sin ({60}^{\circ }-\theta )\sin ({60}^{\circ }+\theta )}{\cos \theta \cos ({60}^{\circ }-\theta )\cos ({60}^{\circ }+\theta )}$

$=\frac{\frac{1}{4}\sin 3\theta }{\frac{1}{4}\cos 3\theta }=\tan 3\theta$

Put $\theta ={20}^{\circ }$

So $\tan \theta \tan ({60}^{\circ }-\theta )\tan ({60}^{\circ }+\theta )=\tan {20}^{\circ }\tan {40}^{\circ }\tan {80}^{\circ }=\tan 3\times {20}^{\circ }=\tan {60}^{\circ }=\sqrt{3}$