Pick any Orthonormal Frame $X_\alpha$ of $TM$, and choose its dual frame $\theta^\alpha$ of $T^*M$. Introduce a collection of 1-forms using covariant derivatives $$ \nabla X_\alpha = \omega_\alpha^\beta X_\beta $$

Proposition : $$ \omega_\alpha^\beta = - \omega_\beta^\alpha $$ $$ d \theta^\alpha = \theta^\beta \wedge \omega_\beta^\alpha $$ Proof: (1) For any vector field $Y$, we have $$ \la \nabla_Y X_\alpha, X_\beta \ra + \la X_\alpha, \nabla_Y X_\beta \ra = 0. $$ (2) Evaluate on $X_\gamma, X_\sigma$, we get $$ LHS=d \theta^\alpha(X_\gamma, X_\sigma) = X_\gamma(\theta^\alpha (X_\sigma)) - X_\sigma(\theta^\alpha(X_\gamma)) - \theta^{\alpha}([X_\gamma, X_\sigma]) = - \theta^{\alpha}([X_\gamma, X_\sigma])$$ Then, we use $$ [X_\gamma, X_\sigma] = \nabla_{X_\gamma} X_\sigma - \nabla_{X_\sigma} X_\gamma = \omega_\sigma^\beta(X_\gamma) X_\beta - \omega_\gamma^\beta(X_\sigma) X_\beta$$ Hence $$- \theta^{\alpha}([X_\gamma, X_\sigma]) = - \omega_\sigma^\alpha(X_\gamma) + \omega_\gamma^\alpha(X_\sigma)$$ And on RHS $$RHS = \theta^\beta \wedge \omega_\beta^\alpha ( X_\gamma, X_\sigma) = \theta^\beta(X_\gamma)\omega_\beta^\alpha(X_\sigma) - \theta^\beta(X_\sigma)\omega_\beta^\alpha(X_\gamma)= \omega_\gamma^\alpha(X_\sigma) - \omega_\sigma^\alpha(X_\gamma)$$