A problem posed by Ramanujan involving $\sum e^{-5\pi n^2}$

Finally I have managed to prove the identity in question. It appears that the main formula (proved in question) as well as its first corollary are both derived from the transformation formula for theta functions rather than being derived from each other. This means that the issue at hand is not really the corollary of the main result as I was expecting.

Let us write $$a=\vartheta(i), b=\vartheta(5i),c=\vartheta(i/5)\tag{1}$$ and we have to prove that $$a=b\sqrt {5\sqrt{5}-10}\tag{2}$$ This is done in two steps and the first one out of these two is obvious. Putting $\tau=5i$ in the transformation formula for theta functions (see the question) we get $$c=b\sqrt{5}\tag{3}$$ In order to prove $(2)$ we need another relation between $a, b$ and $c$ and use it together with $(3)$.

This is the second step where we put $\tau=i+2$ so that $$(-i\tau) ^{-1/2}=\frac{\sqrt{1+2i}}{\sqrt{5}}=\sqrt{\frac{\sqrt{5}+1}{10}}+i\sqrt{\frac{\sqrt{5}-1}{10}}=p+iq\text{ (say)} \tag{4}$$ and $$-\frac{1}{\tau}=\frac{i-2}{5}$$ Using these values in the transformation formula for theta function (and also noting that $\vartheta(\tau +2)=\vartheta(\tau)$) we get $$a=(p+iq)\left\{1+2\sum_{n=1}^{\infty} e^{-\pi n^2/5}\left(\cos\frac{2\pi n^2}{5}-i\sin\frac{2\pi n^2}{5}\right)\right\}$$ Note that the left hand side is purely real and hence equating real parts we get $$a=p\left(1 +2\sum_{n=1}^{\infty} e^{-\pi n^2/5}\cos\frac{2\pi n^2}{5}\right)+2q\sum_{n=1}^{\infty}e^{-\pi n^2/5}\sin\frac{2\pi n^2}{5}$$ and equating imaginary parts we get $$ 2p\sum_{n=1}^{\infty} e^{-\pi n^2/5}\sin\frac{2\pi n^2}{5}=q\left(1+2\sum_{n=1}^{\infty} e^{-\pi n^2/5}\cos\frac{2\pi n^2}{5}\right)$$ Combining these equations we have $$a=\frac{p^2+q^2}{p}\left(1+2\sum_{n=1}^{\infty}e^{-\pi n^2/5}\cos\frac{2\pi n^2}{5}\right)$$ And now we have the magic happening here. If $5\mid n$ then the cosine term equals $1$ otherwise it equals $\cos(2\pi/5)$. We can thus rewrite the above equation as $$a=\frac{p^2+q^2}{p}\left(1+2\cos\frac{2\pi}{5}\sum_{n>0,5\nmid n} e^{-\pi n^2/5}+2\sum_{n=1}^{\infty} e^{-5\pi n^2}\right)$$ and this can be further rewritten as $$a=\frac{p^2+q^2}{p}\left\{1+2\cos\frac{2\pi}{5}\sum_{n=1}^{\infty} e^{-\pi n^2/5}+2\left(1-\cos\frac{2\pi}{5}\right)\sum_{n=1}^{\infty}e^{-5\pi n^2}\right\}$$ Finally this means that \begin{align} a&=\frac{p^2+q^2}{p}\left(1+(c-1)\cos\frac{2\pi}{5}+2(b-1)\sin^2\frac{\pi}{5}\right)\notag\\ &=b\cdot\frac{p^2+q^2}{p}\left(\sqrt{5}\cos\frac{2\pi}{5}+1-\cos\frac{2\pi}{5}\right)\text{ (using (3))}\notag\\ &=\frac{b} {p\sqrt{5}}\left(1+\frac{(\sqrt{5}-1)^2}{4}\right)\text{ (using (4))}\notag\\ &=\frac{b}{p}\cdot\frac{\sqrt{5}-1}{2}\notag\\ &=b\sqrt{\frac{5(\sqrt{5}-1)(3-\sqrt{5})}{4}}\notag\\ &=b\sqrt{5(\sqrt{5}-2)}\notag \end{align} I think this is almost what Ramanujan had in his mind when he posed the problem.