[Math] Prove that $\det(A+B)\det(A-B)=0$

Question

Let $A,B$ be two $n\times n$ matrices with real entries, where $n$ is odd, such that $A\cdot A^{t}=I_n$ and $B\cdot B^{t}=I_n$. Prove that $$\det(A+B)\det(A-B)=0$$
It is obvious that $A^{-1}=A^{t}$ and $B^{-1}=B^{t}$, so $\det A, \det B = \pm 1$. Then I tried to write $\det(A+B)\det(A-B)=\det(A^2-AB+BA-B^2)$, but I didn't get anything useful.

Answer 1

$$\begin{aligned}\det(A+B)\det(A-B)&=\det(A+B)\det(A^T-B^T) \\ &= \det(AA^T+BA^T-AB^T-BB^T)\\&=\det(BA^T-AB^T)\end{aligned}$$

Now use the assumption that $n$ is odd and the fact that $C=BA^T-AB^T$ is skew-symmetric: $C=-C^T$.