Rogers-Ramanujan continued fraction $R(e^{-2 \pi \sqrt 5})$

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).

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.