The ratio of jacobi theta functions Let $q=e^{2\pi i\tau}$. If $\theta_2$ and $\theta_3$ are jacobi theta functions , is it true that the ratio of the two functions can be expressed as a continued fraction of the form
$$
\frac{\theta_2(q^2)}{\theta_3(q^2)}=2q^{1/2}\prod_{n=1}^\infty \frac{(1+q^{4n})^2}{(1+q^{4n-2})^2}=\cfrac{2q^{1/2}}{1-q+\cfrac{q(1+q)^2}{1-q^3+\cfrac{q^2(1+q^2)^2}{1-q^5+\cfrac{q^3(1+q^3)^2}{1-q^7+\ddots}}}}
$$
for $|q|\lt 1$? 
 A: The answer is yes. Given the nome $q = \exp(i\pi\tau)$,  elliptic lambda function $\lambda(\tau)$, Dedekind eta function $\eta(\tau)$, Jacobi theta functions $\vartheta_n(0,q)$, and Ramanujan's octic cfrac, the following relations are known, 
$$\begin{aligned}
u(\tau) & = \big(\lambda(\tau)\big)^{1/8} = \frac{\sqrt{2}\, \eta(\tfrac{\tau}{2})\, \eta^2(2\tau)}{\eta^3(\tau)} = \left(\frac{\vartheta_2(0, q)}{\vartheta_3(0, q)}\right)^{1/2}\\ & = \cfrac{\sqrt{2}\, q^{1/8}}{1 + \cfrac{q}{1 + q + \cfrac{q^2}{1 + q^2 + \cfrac{q^3}{1 + q^3 + \ddots}}}}
\end{aligned}\tag1$$
If we define the cfrac, $$W(q) = \cfrac{1}{1 - q + \cfrac{q(1 + q)^2}{1 - q^3 + \cfrac{q^2(1 + q^2)^2}{1 - q^5 + \cfrac{q^3(1 + q^3)^2}{1 - q^7 + \ddots}}}}$$
we get the $q$-series, 
$$W(q) = 1 - 2q^2 + 5q^4 - 10q^6 + 18q^8 - 32q^{10} + \dots$$
and which is defined in A079006 (after scaling) as the expansion of,
$$W(q) = \frac{1}{q^{1/2}}\left(\frac{\eta(q^2)\, \eta^2(q^8)}{\eta^3(q^4)}\right)^2 = \frac{1}{q^{1/2}}\left(\frac{\eta(2\tau)\, \eta^2(8\tau)}{\eta^3(4\tau)}\right)^2 $$
in powers of $q$. From $(1)$, and since the definition of the Dedekind eta uses the square of the nome as $q = e^{2\pi i \tau}$, we get,
$$\frac{\sqrt{2}\,\eta(2\tau)\, \eta^2(8\tau)}{\eta^3(4\tau)} = \left(\frac{\vartheta_2(0, q^2)}{\vartheta_3(0, q^2)}\right)^{1/2}$$
With basic algebraic substitutions, one then finds that,
$$W(q) = \frac{1}{2q^{1/2}}\frac{\vartheta_2(0, q^2)}{\vartheta_3(0, q^2)}$$
which is exactly what the OP wished to prove. (QED.)
(In fact, Michael Somos in a Sept 2005 comment in the same OEIS link already gave the same cfrac with $q = x^2$.)
(Some more background for those interested.) 
As was pointed out, $\lambda(\tau)$ obeys modular equations. For ex, if $u = \big(\lambda(\tau)\big)^{1/8}$ and $v = \big(\lambda(5\tau)\big)^{1/8}$, then,
$$\Omega_5(u,v) :=u^6 - v^6 + 5u^2 v^2(u^2 - v^2) + 4u v(u^4 v^4 - 1)=0$$
Because of $\Omega_5$, these functions can be used to solve the general quintic, as partly described in this post. And if $k = \lambda(\tau)$ and $l = \lambda(7\tau)$, then,
$$\Omega_7 := (kl)^{\color{red}{1/8}} + \big((1-k)(1-l)\big)^{\color{red}{1/8}} = 1$$
correcting a typo in the Mathworld link with the exponent. Also, $k=\lambda(\sqrt{-n})$, computed in Mathematica as ModularLambda[Sqrt[-n]], is important since it solves the equation,
$$\frac{K'(k)}{K(k)} = \frac{\text{EllipticK[ 1 - ModularLambda[Sqrt[-n]] ]}}{\text{EllipticK[ ModularLambda[Sqrt[-n]] ]}} =\sqrt{n}$$
where $K(k)$ is the complete elliptic integral of the first kind. For example, in his second letter to Hardy, Ramanujan gave the brilliant solution when $n=210$ as,
$$k = \lambda(\sqrt{-210}) = ab \approx 2.706\times 10^{-19}$$
where,
$$a =\big((\sqrt{15}-\sqrt{14})(8-3\sqrt{7})(2-\sqrt{3})(6-\sqrt{35})\big)^2$$
$$b =\big((1-\sqrt{2})(3-\sqrt{10})(4-\sqrt{15})(\sqrt{7}-\sqrt{6})\big)^4$$
So we have this beautiful evaluation of the continued fraction,
$$(ab)^{1/8} = \cfrac{\sqrt{2}\, q^{1/8}}{1 + \cfrac{q}{1 + q + \cfrac{q^2}{1 + q^2  + \ddots}}}$$
when $q = e^{-\pi\sqrt{210}}$. 
