TITLE: Why does $\tan(30^{\large\circ})=\frac{\tan(10^{\large\circ})\tan(50^{\large\circ})}{\tan(20^{\large\circ})}$?
QUESTION [11 upvotes]: This problem is based on this Facebook post.
One can find the value of $x$ in this diagram
 
by noticing that $\angle CBD=50^{\large\circ}$, and therefore,
$$
\frac{\tan\left(10^{\large\circ}\right)}{\tan\left(20^{\large\circ}\right)}
=\frac{ED}{CD}=\frac{\tan(x)}{\tan\left(50^{\large\circ}\right)}\tag1
$$
Solving equation $(1)$ gives
$$
\tan(x)=\frac{\tan\left(10^{\large\circ}\right)\tan\left(50^{\large\circ}\right)}{\tan\left(20^{\large\circ}\right)}\tag2
$$
Numerically computing the arctangent of the quantity in $(2)$ gives $x=30^{\large\circ}$. This surprised me; I had expected some odd angle, but apparently, this turns out to be a nice angle.

My question is: why does $\tan\left(30^{\large\circ}\right)=\frac{\tan\left(10^{\large\circ}\right)\tan\left(50^{\large\circ}\right)}{\tan\left(20^{\large\circ}\right)}$ ?

REPLY [3 votes]: \begin{align}
\frac{\tan10\tan50}{\tan20}
&= \frac{\sin10\sin50\cos20}{\cos10\cos50\sin20}
= \frac{\sin50\cos20}{2\cos^210\cos50}\\
&= \frac{2\cos40\cos20}{2\cos50(1+\cos20)}
= \frac{\cos20+\frac12}{2\cos50+\cos70+\frac{\sqrt3}2}\\
&= \frac{\cos20+\frac12}{\cos50+\cos10+\frac{\sqrt3}2}
= \frac{\cos20+\frac12}{\sqrt3\cos20+\frac{\sqrt3}2}\\
&= \frac1{\sqrt3}=\tan30
\end{align}