In this MSE post, Nicco Mnisi defined a particular $q$-continued fraction of order $12$. More generally, define the cfrac found in Ramanujan's Notebooks, Vol III, Chap. 16, page 24,

$$U(q) = \dfrac{1} {1-ab+\dfrac{(a-bq)(b-aq)} {(1-ab)(1+q^2)+\dfrac{(a-bq^3)(b-aq^3)} {(1-ab)(1+q^4)+\dfrac{(a-bq^5)(b-aq^5)} {(1-ab)(1+q^6)+\ddots }}}}$$

I extrapolated that the general form of Nicco's cfrac apparently is,

$$V(q) = \dfrac{1} {1+ab-\dfrac{(a+bq)(b+aq)} {1+(ab)^3+\dfrac{(a-bq^2)(b-aq^2)q} {1+(ab)^5-\dfrac{(a+bq^3)(b+aq^3)q^2} {1+(ab)^7+\dfrac{(a-bq^4)(b-aq^4)q^3} {(1+(ab)^9-\ddots }}}}} $$

Question: If $ab=q$, and $|q|<1$, is it true that $U(q) = V(q)$?

I tested these numerically for various $a,b,q$ and it seems to be true with $V(q)$ converging faster. However, a rigorous proof is needed.

P.S. With the appropriate factor $q^\alpha(1-q^\beta)$ affixed, these cfracs are algebraic numbers. For example, in this post,

$$\frac{1}{N(e^{-2\pi})}+N(e^{-2\pi}) = \frac{4}{1-\sqrt{3\big(3+\sqrt{3}-3^{3/4}\sqrt{2+\sqrt{3}}\big)}}=536.4953904\dots$$