Let $(M,g)$ be be a Riemannian manifold of dimension 2. Prove that if $E_1,E_2$, $\bar{E_1},\bar{E_2}$ are two different frames, then $\omega_1^{2}=\bar\omega_1^{2}+d\varphi$, where $\omega_1^2$ is the connection form on $E_1,E_2$, $\bar\omega_1^2$ is the connection form on $\bar{E_1},\bar{E_2}$ and $\varphi$ is the $0$-form (smooth function) that for each $p\in M$ returns the angle between $E_1(p)$ and $\bar {E_1}(p)$.
My attempt:
I was able to prove that since $E_1,E_2$ is an orthonormal basis (for each $p\in M$) for $T_pM$, then $\bar{E_1}=cos(\varphi)E_1+sin(\varphi)E_2$ and $\bar{E_2}=-sin(\varphi)E_1+cos(\varphi)E_2$, but I don't know how to reach the final conclusion.
Edit:
$\omega_1^{2}=d\theta_1(E_1,E_2)\theta_1+d\theta_2(E_1,E_2)\theta_2\\ \bar{\omega}_1^{2}=d\bar{\theta_1}(\bar{E_1},\bar{E_2})\bar{\theta_1}+d\bar{\theta_2}(\bar{E_1},\bar{E_2})\bar{\theta_2}$
Where $\theta_1,\theta_2,\bar\theta_2,\bar\theta_1$ are the corresponding dual basis (depending on $p\in M$).
Any help would be appreciated.