If ${A}_{ab} = \delta_{ab} + \varepsilon_{abc}n^c$ and $B^{ab} = \frac{1}{1+n^2}(\delta^{ab} + n^an^b - \varepsilon^{abc}n_c)$ what is the correct way to evaluate $$C^{ab} = (AB)^{ab} $$
Here, $\delta_{ab}$ is the Kronecker symbol, $n$ is a vector and $\varepsilon$ is the totally anti-symmetric unit tensor.
Presumably, the point of this exercise was for you utilize the properties of the Levi-Civita tensor $\varepsilon$ to discover that $$ C^{ab} = \delta^{ab} $$ For convenience, let $N = \varepsilon \cdot n$ which is a skew-symmetric second order tensor, and $s = 1 + n^2$ which is a scalar. Then $$ \eqalign { A &= I + N \cr sB &= I + nn - N \cr } $$ Their product is $$ \eqalign { sA\cdot B &= (I + N)\cdot(I + nn - N) \cr &= (I + nn - N) + (N + N\cdot nn - N^2) \cr &= I + nn - N^2 \cr &= I + nn - (nn-n^2 I) \cr &= I + n^2 I \cr &= sI \cr } $$ In the preceding, I used the fact that $N\cdot n = \varepsilon : nn = 0$, since $nn$ is symmetric and $\varepsilon$ is skew in all indices. I'll leave it to you to confirm that $N^2 = (nn-n^2I)$ for which this page should be helpful.