How to find the numerical value of $\tan20\tan40\tan60\tan80$?

Question

$$\tan20\tan40\tan60\tan80$$

I tried to bring $\tan20$, $\tan40$ in $\tan(a+b)$ form but couldn't get the answer. Using calculator I got it has $3$. Can anybody please tell me how to solve this?

Answer 1

$$\tan{x}\tan(x+60^{\circ})\tan(x+120^{\circ})=\tan{x}\cdot\tfrac{\tan{x}+\sqrt3}{1-\sqrt3\tan{x}}\cdot\tfrac{\tan{x}-\sqrt3}{1+\sqrt3\tan{x}}=\tfrac{\tan^3x-3\tan{x}}{1-3\tan^2x}=-\tan3x.$$ Thus, $$\tan20^{\circ}\tan40^{\circ}\tan60^{\circ}\tan80^{\circ}=-\tan20^{\circ}\tan80^{\circ}\tan140^{\circ}\tan60^{\circ}=-(-\sqrt3)\sqrt3=3.$$

Answer 2

In general, $$\tan x=\frac{\sin x}{\cos x}=\frac{\sin^2x}{\sin x\cos x} =\frac{2\sin^2x}{\sin 2x}.$$ Using this, will cause your product to telescope: \begin{align} \tan20^\circ\tan40^\circ\tan60^\circ\tan80^\circ &=16\frac{\sin^220^\circ\sin^240^\circ\sin^260^\circ\sin^280^\circ} {\sin40^\circ\sin80^\circ\sin120^\circ\sin160^\circ}\\ &=16\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ=16P \end{align} say. Now \begin{align} P^2&=\prod_{k=1}^8\sin(20k)^\circ=\frac1{2^8}(e^{-ik\pi/9}-e^{ik\pi/9}) =\frac1{2^8}(1-e^{i2k\pi/9}). \end{align} But $$\frac{X^9-1}{X-1}=\prod_{k=1}^8(X-e^{i2k\pi/9})$$ and taking $X=1$ gives $P^2=9/2^8$. So $P=3/16$ and your product is $3$.