In the book of large deviations from den Hollander, in page 77 one reads
where $\rho_0 = \frac{1-\omega_0}{\omega_0}$ and equation VII.14 is
However, combining these two I have arrived at
$$\frac{\varphi(r,\omega)}{\tilde{\varphi}(r,\omega)} = \frac{\omega_0 e^ r + (1 - \omega_0)e^r \varphi(r,\sigma^{-1}(\omega)) \varphi(r,\omega)}{(1 - \omega_0) e^ r + \omega_0e^r \tilde{\varphi}(r,\sigma(\omega)) \tilde{\varphi}(r,\omega)} $$
dividing by $\omega_0 e^r$ numerator and denominator we arrive at:
$$\frac{\varphi(r,\omega)}{\tilde{\varphi}(r,\omega)} = \frac{1 + \rho_0 \varphi(r,\sigma^{-1}(\omega)) \varphi(r,\omega)}{\rho_0+ \tilde{\varphi}(r,\sigma(\omega)) \tilde{\varphi}(r,\omega)} = q $$
So now we write $\varphi(r,\omega) = q \tilde{\varphi}(r,\omega)$ to obtain:
$$\frac{\rho_0\varphi(r,\omega)}{\tilde{\varphi}(r,\omega)} = \frac{1 + \rho_0 \varphi(r,\sigma^{-1}(\omega)) \varphi(r,\omega)}{1 + (\rho_0)^{-1} \tilde{\varphi}(r,\sigma(\omega)) \tilde{\varphi}(r,\omega)} \\ =\frac{1 + \rho_0q \varphi(r,\sigma^{-1}(\omega)) \tilde{\varphi}(r,\omega)}{1 + (\rho_0q)^{-1} \tilde{\varphi}(r,\sigma(\omega)) \varphi(r,\omega)} $$
But how to arrive at the desired expression?