In local inertial coordinates $g_{\mu \nu}$ becomes $\eta_{\mu \nu}$. Also, tetrads are defined as the basis vectors $ {\hat{e}}_{a}, {\hat{e}}_{b}$... such that $g( {\hat{e}}_{a},{\hat{e}}_{b} ) = \eta_{ab}$. So, this means that in tetrad basis $g_{\mu \nu} = \eta_{\mu \nu}$. Hence the bases of local inertial coordinates should be the same as the tetrad bases (apart from a Lorentz transformation). Is that correct?
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$\begingroup$ In inertial coordinates (in general) the metric is $\eta_{\mu\nu}$ only at one point. $\endgroup$– MBNCommented Nov 12, 2017 at 10:28
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Yes, that is correct. If you Lorentz transform a tetrad, $\eta_{\mu\nu}$ is preserved by the transformation.
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$\begingroup$ I think it is correct after having gone through Steven Weinberg's 'Gravitation and Cosmlogy' (section 4.5: The Tetrad Formalism). $\endgroup$ Commented Nov 12, 2017 at 5:03