Evaluating $\left| \frac{\tan40^\circ + \tan100^\circ + \tan160^\circ}{\tan20^\circ\tan40^\circ\tan80^\circ} \right| $ 
How do I find the value of the following expression?
$$
\left| \frac{\tan40^\circ + \tan100^\circ + \tan160^\circ}{\tan20^\circ\tan40^\circ\tan80^\circ} \right|
$$

I tried writing the numerator as $\tan 40^\circ - \tan80^\circ -\tan20^\circ,$ but then the expression was getting complicated.
 A: First of all we have (see Morrie's law)
$$
\tan20^\circ\tan40^\circ\tan80^\circ=\sqrt{3}.
$$
The numerator is, setting $x=20^\circ,$
\begin{align}
&\tan40^\circ+\tan100^\circ+\tan160^\circ=\\
&\qquad\qquad=\tan(60^\circ-x)+\tan(120^\circ-x)+\tan(180^\circ-x)=\\
&\qquad\qquad=
  \frac{\tan 60^\circ-\tan x}{1+\tan 60^\circ\tan x}+
  \frac{\tan120^\circ-\tan x}{1+\tan120^\circ\tan x}+
  \frac{\tan180^\circ-\tan x}{1+\tan180^\circ\tan x}=\\
&\qquad\qquad=
  \frac{ \sqrt{3}-\tan x}{1+\sqrt{3}\tan x}+
  \frac{-\sqrt{3}-\tan x}{1-\sqrt{3}\tan x}-
  \tan x=\\
&\qquad\qquad=
  \frac{\sqrt{3}\cos x-\sin x}{\cos x+\sqrt{3}\sin x}-
  \frac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x}-
  \frac{\sin x}{\cos x}=\\
&\qquad\qquad=
  -3\cdot\frac{3\sin x\cos^2 x-\sin^3 x}{\cos^3 x-3\sin^2 x\cos x}=\\
&\qquad\qquad=-3\cdot\frac{\sin(3x)}{\cos(3x)}=-3\tan60^\circ=-3\sqrt{3}
\end{align}
So the final result is
$$
\left| \frac{\tan40^\circ + \tan100^\circ + \tan160^\circ}{\tan20^\circ\tan40^\circ\tan80^\circ} \right|=\left|\frac{-3\sqrt{3}}{\sqrt{3}}\right|=3
$$
A: Notice that each $\newcommand{\degree}{{\lower{.5pt}\Large\circ}}x\in\left\{20^\degree,-40^\degree,80^\degree\right\}$ satisfies
$$
\begin{align}
\sqrt3
&=\tan(3x)\\
&=\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}\tag1
\end{align}
$$
Thus,
$$
\tan^3(x)-3\sqrt3\tan^2(x)-3\tan(x)+\sqrt3=0\tag2
$$
Vieta says that the sum of the roots is the negative of the coefficient of $\tan^2(x)$. That is,
$$
\tan\left(20^\degree\right)-\tan\left(40^\degree\right)+\tan\left(80^\degree\right)=3\sqrt3\tag3
$$
and the product of the roots is the negative of the constant term. That is,
$$
-\tan\left(20^\degree\right)\tan\left(40^\degree\right)\tan\left(80^\degree\right)=-\sqrt3\tag4
$$
Therefore,
$$
\begin{align}
\frac{\tan\left(40^\degree\right)+\tan\left(100^\degree\right)+\tan\left(160^\degree\right)}{\tan\left(20^\degree\right)\tan\left(40^\degree\right)\tan\left(80^\degree\right)}
&=\frac{\tan\left(40^\degree\right)-\tan\left(80^\degree\right)-\tan\left(20^\degree\right)}{\tan\left(20^\degree\right)\tan\left(40^\degree\right)\tan\left(80^\degree\right)}\\[6pt]
&=-3\tag5
\end{align}
$$
Just take the absolute value of $(5)$.
