Why does $\tan(30^{\large\circ})=\frac{\tan(10^{\large\circ})\tan(50^{\large\circ})}{\tan(20^{\large\circ})}$? 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 $\tan\left(30^{\large\circ}\right)=\frac{\tan\left(10^{\large\circ}\right)\tan\left(50^{\large\circ}\right)}{\tan\left(20^{\large\circ}\right)}$ ?

 A: \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}
A: The task is to show that $\tan20^\circ\tan30^\circ=\tan10^\circ\tan50^\circ$. After  multiplication through by the cosines of the angles, the task is converted to one of proving that the following quantity is zero:$$a:=\sin20^\circ\sin30^\circ\cos10^\circ\cos50^\circ-\cos20^\circ\cos30^\circ\sin10^\circ\sin50^\circ.$$ Now, from the standard trigonometric formulas $2\cos A\cos B=\cos(A-B)+\cos(A+B)$ and $2\sin A\sin B=\cos(A-B)-\cos(A+B)$, we get  $$2a=(\cos10^\circ-\cos50^\circ)\cos10^\circ\cos50^\circ-(\cos10^\circ+\cos50^\circ)\sin10^\circ\sin50^\circ\qquad\qquad\qquad\quad$$ $$=(\cos10^\circ\cos50^\circ-\sin10^\circ\sin50^\circ)\cos10^\circ-(\cos10^\circ\cos50^\circ+\sin10^\circ\sin50^\circ)\cos50^\circ$$ $$=\cos60^\circ\cos10^\circ-\cos40^\circ\cos50^\circ$$ $$\qquad\qquad=\tfrac12\cos10^\circ-\tfrac12(\cos10^\circ+\cos90^\circ)$$ $$=0.$$
A: Answer to the Question
$$
\begin{align}
\frac{\tan\left(10^{\large\circ}\right)\tan\left(50^{\large\circ}\right)}{\tan\left(20^{\large\circ}\right)}
&=\frac1{\tan\left(20^{\large\circ}\right)}
\overbrace{\frac{\tan\left(30^{\large\circ}\right)-\tan\left(20^{\large\circ}\right)}{1+\tan\left(30^{\large\circ}\right)\tan\left(20^{\large\circ}\right)}}^{\tan\left(10^{\large\circ}\right)}
\overbrace{\frac{\tan\left(30^{\large\circ}\right)+\tan\left(20^{\large\circ}\right)}{1-\tan\left(30^{\large\circ}\right)\tan\left(20^{\large\circ}\right)}}^{\tan\left(50^{\large\circ}\right)}\tag1\\
&=\frac{\frac13-\tan^2\left(20^{\large\circ}\right)}{\tan\left(20^{\large\circ}\right)-\frac13\tan^3\left(20^{\large\circ}\right)}\tag2\\
&=\frac1{\tan\left(60^{\large\circ}\right)}\tag3\\[6pt]
&=\tan\left(30^{\large\circ}\right)\tag4
\end{align}
$$
Explanation:
$(1)$: $\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$
$(2)$: $\tan\left(30^{\large\circ}\right)=\frac1{\sqrt3}$
$(3)$: $\tan(3x)=\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}$
$(4)$: $\frac1{\tan\left(60^{\large\circ}\right)}=\tan\left(90^{\large\circ}-60^{\large\circ}\right)$

General Identity
The answer above is a special case of the identity
$$
\cot(3x)=\frac{\tan\left(\frac\pi6-x\right)\tan\left(\frac\pi6+x\right)}{\tan(x)}\tag5
$$
whose proof mirrors the answer above.
