So I have asked this question recently, and it is essentially the application of the Jacobi Triple Identity as the problem suggest we should use this, however there was an answer posted to the recent thread I made Gauss identities of $q$-analog, but when I tried to study it carefully I still do not get the exact application of Jacobi's identity (I even saw the first equality somewhere on this forum Prove these identities using Jacobi's triple product identity., which seems somewhat satisfactory, however, the second equality I am still very much stuck), but the product terms involving thee $q$'s got me really confused) to get the results I want, which are:
$$\prod_{k\ge1}\frac{1-q^k}{1+q^k}=\sum_{{n\in\mathbb{Z}}}(-1)^nq^{n^2}$$ $$\prod_{k\ge1}\frac{1-q^{2k}}{1-q^{2k-1}}=\sum_{n\ge0}q^{\binom{n+1}{2}}$$
Could anyone give me a clear step-by-step explanation of how to use the Jacobi triple product identity to prove the above results? I would appreciate the help.
From JTP $$\sum_n(-1)^nq^{n^2}=\prod_{n=1}^\infty(1-q^{2n})(1-q^{2n-1})^2 =\prod_{n=1}^\infty(1-q^n)(1-q^{2n-1}) =\prod_{n=1}^\infty\frac{(1-q^n)^2}{1-q^{2n}} =\prod_{n=1}^\infty\frac{(1-q^n)^2}{(1-q^n)(1+q^n)} =\prod_{n=1}^\infty\frac{1-q^n}{1+q^n}. $$ Here I twice used $$\prod_{n=1}^\infty(1-q^{2n})(1-q^{2n-1}) =\prod_{n=1}^\infty(1-q^n).$$
For the second one, write $$\sum_{n=0}^\infty q^{\binom{n+1}2}=\frac12\sum_{n=-\infty}^\infty q^{n/2}q^{n^2/2}$$ and then use JTP.