This identity is solvable by the help of trigonometry identities, but I guess there is an interesting and simple geometry interpretation behind this identity and I can't find it.
I found it when I was thinking about World's Hardest Easy Geometry Problem
Consider a triangle $ABC$ with $\angle A = 10^\circ$, $\angle B = 150^\circ$, and $\angle C=20^\circ$. Let $O$ be the circumcenter of the triangle $ABC$. Then $$\angle AOC = 360^\circ - 2 \angle B = 60^\circ,$$ so triangle $AOC$ is equilateral.
Let $S$ be the circumcenter of the triangle $OBC$. Then $$\angle BSC = 2\angle BOC = 4\angle BAC = 40^\circ$$ and $$\angle SCB = \frac{180^\circ - \angle BSC}2 = 70^\circ,$$ so $\angle SCA = 50^\circ$. Denoting the intersection of $AC$ and $SB$ by $X$ we get $$\angle CXS = 180^\circ - \angle SCA - \angle BSC = 90^\circ$$ so $AC \perp SB$.
Moreover $\triangle ASC \equiv \triangle ASO$ because $AC=AO$ and $SC=SO$. In particular $\angle CAS = \angle SAO$ and since $\angle CAO=60^\circ$, we have $\angle CAS = 30^\circ$.
Observe that $$\frac{\tan \angle BAC}{\tan \angle ACB} = \frac{\frac{BX}{AX}}{\frac{BX}{CX}} = \frac{CX}{AX} = \frac{\frac{SX}{AX}}{\frac{SX}{CX}} = \frac{\tan \angle CAS}{\tan \angle SCA},$$ therefore $$\frac{\tan 10^\circ}{\tan 20^\circ} = \frac{\tan 30^\circ}{\tan 50^\circ}.$$
