Rogers-Ramanujan continued fraction $R(e^{-2 \pi \sqrt 5})$ Let  $$R(q) = \cfrac{q^{1/5}}{1 + \cfrac{q}{1 + \cfrac{q^{2}}{1 + \cfrac{q^{3}}{1 + \cdots}}}}$$
It is easy to evaluate $R(e^{-2 \pi/ \sqrt 5})$ using the Dedekind eta function identity $\eta(-\frac{1}{z})=\sqrt{-iz}\eta(z)$   
and  one of the most fundamental properties of $R(q)$ $$\frac{1}{R(q)}-1-R(q)=\frac{f(-q^\frac{1}{5})}{q^\frac{1}{5}f(-q^5)}=\frac{\eta(\tau/5)}{\eta(5\tau)}$$
Where $q=e^{2 \pi i \tau}$ and $f(q)$ is the theta function (Ramanujan's notation). Then
$$R(e^{-2 \pi/ \sqrt 5})=\sqrt[5]{\sqrt{1+\beta^{10}}-\beta^5}$$
where $$\beta=\frac{1+\sqrt{5}}{2 } $$ is the golden ratio.
If $\alpha_1 , \alpha_2>0$ and $\alpha_1 \alpha_2=\pi^2$ then
$$\bigg(\frac{1+\sqrt{5}}{2}+R(e^{-2 \alpha_1})\bigg)\bigg(\frac{1+\sqrt{5}}{2}+R(e^{-2 \alpha_2})\bigg)=\frac{5+\sqrt{5}}{2}$$
using this identity, I evaluate  $R(e^{-2 \pi \sqrt 5})$ Since I already know the value of the $R(e^{- 2 \pi/ \sqrt5})$:
$$\color{blue} {R(e^{-2 \pi \sqrt 5})=  \frac{\beta+2}{\beta+\sqrt[5]{\sqrt{1+\beta^{10}}-\beta^5}}-\beta}.$$
There is another way to evaluate $R(e^{-2 \pi \sqrt 5})$.
$$R(e^{-2 \pi \sqrt 5})=\sqrt{(\frac{A+1}{2})+1}-\frac{A+1}{2}$$
where $A$ satisfies the quadratic equation
$$\frac{A}{\sqrt{5}V}-\frac{\sqrt{5}V}{A}=\bigg(V-V^{-1}\bigg)^2 \bigg(\frac{V-V^{-1}}{\sqrt{5}}-\frac{\sqrt{5}}{V-V^{-1}}\bigg)$$
and
$$V=\frac{G_{125}}{G_5}$$
where $G_n=2^{-1/4}e^{\pi \sqrt{n}/24} \chi(e^{- \pi \sqrt{n}})$ is the Ramanujan's class invariant(algebraic), $\chi(q)$ is the Ramanujan's fuction defined by  $\chi(-q)=(q;q)_\infty$, and $(a;q)_n$ is a q-Pochhammer symbol.
There are my questions:
$1.$ How to calculate the class invariant $G_{125}$ in order to evaluate $R(e^{-2 \pi \sqrt 5})$ as indicated above?
$2.$ Does there exist another way to evaluate $R(e^{-2 \pi \sqrt 5})$ without using $R(e^{-2 \pi / \sqrt 5})$ and class invariants?
 A: The result $$\small R(e^{-2\pi\sqrt{5}})=\frac{\sqrt{5}}{1+\left[5^{3/4}\left(\frac{\sqrt{5}-1}{2}\right)^{5/2}-1\right]^{1/5}}-\frac{\sqrt{5}+1}{2}=\frac{\beta+2}{\beta+\sqrt[5]{\sqrt{1+\beta^{10}}-\beta^5}}-\beta$$ was one of the results of Ramanujan communicated to Hardy in his second letter. It was proved by Watson in 1929 (http://www.inp.nsk.su/~silagadz/Ramanujan_continued_fraction1.pdf) and by  Ramanathan in 1984 (http://www.inp.nsk.su/~silagadz/Ramanujan_continued_fraction2.pdf).
A: Let $R(q)$ be the Rogers-Ramanujan continued fraction
$$
R(q):=\frac{q^{1/5}}{1+}\frac{q^1}{1+}\frac{q^2}{1+}\frac{q^3}{1+}\ldots,|q|<1
$$
Let also for $r>0$ 
$$
Y=Y(r):=R(e^{-2\pi\sqrt{r}})^{-5}-11-R(e^{-2\pi\sqrt{r}})^5
$$
It is easy to show someone that
$$
Y\left(\frac{r}{5}\right)Y\left(\frac{1}{5r}\right)=125, : (1)
$$
for all $r>0$. Hence for $r=1$
$$
Y\left(\frac{1}{5}\right)=\sqrt{125}=5\sqrt{5}
$$ 
hence
$$
R\left(e^{-2\pi/\sqrt{5}}\right)=\sqrt[5]{\frac{2}{11+5\sqrt{5}+\sqrt{250+110\sqrt{5}}}}
$$
Using the modular relation
$$
R\left(e^{-2\pi\sqrt{1/r}}\right)=\frac{-(1+\sqrt{5})R\left(e^{-2\pi\sqrt{r}}\right)+2}{2R\left(e^{-2\pi\sqrt{r}}\right)+1+\sqrt{5}} : (2)
$$
we easily get the result.
Proof of (1):
The fifth degree modular equation of $R(q)$ is
$$
R\left(q^{1/5}\right)^5=R(q)\frac{1-2R(q)+4R(q)^2-3R(q)^3+R(q)^4}{1+3R(q)+4R(q)^2+2R(q)^3+R(q)^4}
$$
Also it holds (2), with $v_r=R\left(e^{-2\pi\sqrt{r}}\right)$, $r>0$ (see: W. Duke. 'Continued Fractions and Modular Functions'):
$$
R\left(e^{-2\pi\sqrt{1/r}}\right)=v_{1/r}=\frac{-(1+\sqrt{5})v_r+2}{2v_r+1+\sqrt{5}}
$$
A routine algebraic evaluation can show us that
$$
\left(v_{r/25}^{-5}-11-v_{r/25}^5\right)\left(v_{1/r}^{-5}-11-v_{1/r}^{5}\right)=\ldots=125
$$
Hence the proof is complete.
