Can anyone guide me how can I prove these two identities?
a)$$\prod_{n=1}^{\infty}\frac{1-q^{2n}}{1-q^{2n-1}}=\sum^{\infty}_{n=1}q^{n(n+1)/2}$$ b) $$\prod_{n=1}^{\infty}\frac{(1+q^{2n})^2(1-q^{2n})^2}{(1+q^{2n-1})^2(1-q^{2n-1})^2}=\sum^{\infty}_{n=0}\frac{(-1)^nq^n}{1-q^{2n+1}}$$
The first equality follows from the Jacoby triple product taked in this form $$ \prod_{n\geq0}\left\{ \left(1+q^{2kn+k-l}\right)\left(1+q^{2kn+k+l}\right)\left(1-q^{2kn+2k}\right)\right\} =\sum_{n=-\infty}^{\infty}q^{kn^{2}+nl}. $$ In fact taking $k=l=1/2 $ we get $$\prod_{n\geq0}\left\{ \left(1+q^{n}\right)\left(1-q^{2n+2}\right)\right\} =\sum_{n=-\infty}^{\infty}q^{n\left(n+1\right)/2} $$ and now we can observe that $$\left(1+q\right)\left(1+q^{2}\right)\left(1+q^{3}\right)\cdots=\frac{\left(1-q^{2}\right)\left(1-q^{4}\right)\left(1-q^{6}\right)}{\left(1-q\right)\left(1-q^{2}\right)\left(1-q^{3}\right)}\cdots=\frac{1}{\left(1-q\right)\left(1-q^{3}\right)\left(1-q^{5}\right)}\cdots $$ and so $$2\prod_{n\geq1}\frac{1-q^{2n}}{1-q^{2n-1}}=\sum_{n=-\infty}^{\infty}q^{n\left(n+1\right)/2}. $$ Now remain to note that $$\sum_{n=-\infty}^{\infty}q^{n\left(n+1\right)/2}=2\sum_{n=0}^{\infty}q^{n\left(n+1\right)/2} $$ and we have done.