Let $t$ be a time function and $t^a$ the time flow vector such that $t^a\nabla_a t=1$. Let $\Sigma_t$ be a hypersurface of constant $t$ with unit normal $n^a$, $n^a n_a=-1$. Wald (1984), p. 255 defines the lapse function as $$N=-t^a n_a=\frac{1}{n^a\nabla_a t}$$ I am seriously stuck on the second equality. I really have no clue how to prove it.
Any help would be greatly appreciated.
Edit: A typo in the original question has been corrected.
$n^a = (-\nabla_b t \nabla^b t)^{-1/2}\nabla^a t$ since $n^a$ is hypersurface orthogonal to the foliation $\Sigma_t$ and unit timelike, therefore $-t^a n_a = -(-\nabla_b t \nabla^b t)^{-1/2}$ and $n_a n^a = -1$ which gives $-(-\nabla_b t \nabla^b t)^{-1/2} = (n_a \nabla^a t)^{-1}$ hence $-t^a n_a =(n_a \nabla^a t)^{-1}$ as desired.
I think the discussion above is overcomplicating this. The covectors $n_a$ and $\nabla_a t$ both have zero contraction with any vector tangent to the constant $t$ hypersurface, so they must be proportional: $n_a = -N \nabla_a t$ for some function $N$. Contracting both sides with $n^a$ yields $N=1/(n^a\nabla_a t)$. Contracting both sides with $t^a$ yields $t^a n_a=-N$. (Note the typo in the question: it should read $t^a\nabla_a t = 1$ in the first line.)
