The Rogers-Ramanujan cfrac is,
$$r = r(\tau)= \cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\ddots}}}$$
If $q = \exp(2\pi i \tau)$, then it is known that,
$$\frac{1}{r}-r =\frac{\eta(\tau/5)}{\eta(5\tau)}+1\tag1$$
$$\frac{1}{r^5}-r^5 =\left(\frac{\eta(\tau)}{\eta(5\tau)}\right)^6+11\tag2$$
with the Dedekind eta function, $\eta(\tau)$.
Q: Is there a similar simple identity known using ratios of the Jacobi theta functions $\vartheta_n(0,q)$?
$\color{brown}{Edit}$: In response to a comment, here are some details. Suppose we don't know $(2)$. One way to find such relations is to use known identities. Given the j-function $j(\tau)$, we have,
$$j(\tau)=-\frac{(r^{20} - 228r^{15} + 494r^{10} + 228r^5 + 1)^3}{r^5(r^{10} + 11r^5 - 1)^5}\tag3$$
$$j(\tau)=\frac{(5x^2+10x+1)^3}{x^5}\tag4$$
where $x = \left(\frac{\eta(\tau)}{\sqrt{5}\,\eta(5\tau)}\right)^6$. Equate $(3),\,(4)$,
$$-\frac{(r^{20} - 228r^{15} + 494r^{10} + 228r^5 + 1)^3}{r^5(r^{10} + 11r^5 - 1)^5} = \frac{(5x^2+10x+1)^3}{x^5}\tag5$$
and using symbolic software like Mathematica to factor $(5)$, we find one factor is given by,
$$\frac{1}{r^5}-r^5 =125x+11\tag6$$
and it only takes minor tweaking to make $(6)$ have the form of $(2)$. Thus, all we need is to express $j(\tau)$ not by an eta quotient $x$, but by a theta quotient $y=\left(\frac{\vartheta_n(0,q)}{\vartheta_n(0,q^5)}\right)^k$ so that,
$$j(\tau) = \frac{f_1(y)}{f_2(y)}\tag7$$
is a ratio of polynomials in $y$.
$\color{brown}{P.S.}$ The motivation for this is that the general quintic is solvable by the Rogers-Ramanujan cfrac via the eta quotient above as described in this post.
Motivated by ccorns's comment, I decided to re-visit this question. After some effort, I found what I was looking for. The Rogers-Ramanujan cfrac is,
$$r = r(\tau)= \cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\ddots}}}$$
where $q = \exp(2\pi i \tau)$. It turns out we have the nice relations,
$$\begin{aligned}\frac{1}{r^5}-r^5 &=\,u^3+11\\ &= \frac{v(v-5)^2}{(v-1)^2}+11 \end{aligned}$$
where the three roots of the cubic in $u$ are eta quotients,
$$u_n = \left(\frac{\eta(\tau')}{\eta(5\tau')}\right)^2$$
with $\tau'=\tau+n$, while the three roots of the cubic in $v$ are theta quotients,
$$v_n = \left(\frac{\vartheta_{n+2}(0,p)}{\vartheta_{n+2}(0,p^5)}\right)^2$$
for the nome $p = e^{\pi i \tau}$ and for $n=0,1,2$.