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