a conjecture of certain q-continued fractions Given the squared nome $q=e^{2i\pi\tau}$ with $|q|\lt1$, define,
$$\begin{aligned}F(q)=\cfrac{1-q^2}{1-q^3+\cfrac{q^3(1-q)(1-q^5)}{1-q^9+\cfrac{q^6(1-q^4)(1-q^8)}{1-q^{15}+\cfrac{q^9(1-q^7)(1-q^{11})}{1-q^{21}+\ddots}}}}\overset{\color{red}{?}}=\prod_{n=1}^\infty\frac{(1-q^{12n-2})(1-q^{12n-10})}{(1-q^{12n-4})(1-q^{12n-8})}\\[1.5mm]&\end{aligned}$$
and
$$\begin{aligned}G(q)=\cfrac{1-q^4}{1-q^3+\cfrac{q^3(1-\frac{1}{q})(1-q^7)}{1-q^9+\cfrac{q^6(1-q^2)(1-q^{10})}{1-q^{15}+\cfrac{q^9(1-q^5)(1-q^{13})}{1-q^{21}+\ddots}}}}\overset{\color{red}{?}}=\prod_{n=1}^\infty\frac{(1-q^{12n-4})(1-q^{12n-8})}{(1-q^{12n-2})(1-q^{12n-10})}\\[1.5mm]&\end{aligned}$$
Q: How do we prove rigorously that the two q-continued fractions are equal to the q-series?
(if true, then the two q-continued fractions are reciprocals such that their product is unity, $F(q)\,G(q)=1$.)
 A: (A half-answer.) Let $|q|<1$ and,
$$p = q^m,\quad \alpha = \pm q^{-n},\quad \beta = \pm q^n,\quad \alpha\beta = 1$$ 
We propose that,
$$\small C_{m,n}(q)=\prod_{k=1}^\infty\frac{(1-\alpha\, p^{4k})(1-\beta\, p^{4k-4})}{(1-\alpha\, p^{4k-2})(1-\beta\, p^{4k-2})} \overset{\color{red}{?}}=\cfrac{(1-\beta\,p^0)}{ 1-p+\cfrac{p(1-\alpha\, p)(1-\beta\, p)}{1-p^3+\cfrac{p^2(1-\alpha\, p^2)(1-\beta\, p^2)}{1-p^{5}+\cfrac{p^3(1-\alpha\, p^3)(1-\beta\, p^3)}{1-p^{7}+\ddots}}}}\tag1 $$
It takes only some algebraic manipulation to show that given,
$$\alpha_1 = \pm q^{\large -n_1},\quad \beta_1 = \pm q^{\large n_1},\quad \alpha\beta = 1$$
then the two products are reciprocals,
$$\frac{(1-\alpha_1\, p^{4k})(1-\beta_1\, p^{4k-4})}{(1-\alpha_1\, p^{4k-2})(1-\beta_1\, p^{4k-2})} \times \frac{(1-\alpha_2\, p^{4k})(1-\beta_2\, p^{4k-4})}{(1-\alpha_2\, p^{4k-2})(1-\beta_2\, p^{4k-2})}=1$$
if,
$$m =\frac{n_1+n_2}{2}\tag2$$ 
For example, let $m,\,n_1,\,n_2 = 3,\,2,\, 4$, and using $(1)$ and the positive case of $\pm$, we get,
$$C_{3,2}(q) = \prod_{n=1}^\infty\frac{(1-q^{12n-2})(1-q^{12n-10})}{(1-q^{12n-4})(1-q^{12n-8})} = \small\cfrac{1-q^2}{1-q^3+\cfrac{q^3(1-q)(1-q^5)}{1-q^9+\cfrac{q^6(1-q^4)(1-q^8)}{1-q^{15}+\cfrac{q^9(1-q^7)(1-q^{11})}{1-q^{21}+\ddots}}}}$$
$$C_{3,4}(q) = \prod_{n=1}^\infty\frac{(1-q^{12n-4})(1-q^{12n-8})}{(1-q^{12n-2})(1-q^{12n-10})} = \small\cfrac{1-q^4}{1-q^3+\cfrac{q^3(1-q^{-1})(1-q^7)}{1-q^9+\cfrac{q^6(1-q^2)(1-q^{10})}{1-q^{15}+\cfrac{q^9(1-q^5)(1-q^{13})}{1+q^{21}+\ddots}}}}$$
the same as in the post and,
$$C_{3,2}(q)\,C_{3,4}(q) = 1$$
The reciprocality is true for general $m,n_1,n_2$ that obey $(2)$. However, what remains is to show that the cfrac is indeed equal to the infinite product $(1)$. 
