Given $q=e^{2i\pi\tau}$ with $|q|\lt1$, define the well-known Göllnitz-Gordon identities $$A(q)=\sum_{n=0}^\infty \frac{q^{n(n+1)}(-q;q^2)_n}{(q^2;q^2)_n}=\prod_{n=1}^\infty \frac{1}{(1-q^{8n-3})(1-q^{8n-4})(1-q^{8n-5})}, \\ \phantom{a} \\ B(q)=\sum_{n=0}^\infty \frac{q^{n^2}(-q;q^2)_n}{(q^2;q^2)_n}=\prod_{n=1}^\infty \frac{1}{(1-q^{8n-1})(1-q^{8n-4})(1-q^{8n-7})},$$ as well as the following conjectured $q$-continued fraction, which is related to the Göllnitz-Gordon identities:
$$H(q)=\cfrac{q^{1/2}(1-q)}{1+q^2-\cfrac{q^2(1+q)(1+q^3)}{1+q^6+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^{10}-\cfrac{q^6(1+q^5)(1+q^7)}{1+q^{14}+\cfrac{q^8(1-q^7)(1-q^9)}{1+q^{18}-\ddots}}}}}$$
One can analogously compare it with the famous Rogers–Ramanujan continued fraction which is related to the Rogers–Ramanujan identities.
Question: How do we show that $$H(q) = \frac{q^{1/2}A(q)}{B(q)}\text{?}$$