For orthogonal matrices $A$ and $B$, prove $\det(A^{t}B - B^{t}A)=\det(A+B)\det(A-B)$

Question

I can't prove this formula: $$ \det(A^{t}B - B^{t}A)=\det(A+B)\det(A-B) $$

I tried using fact that $A^{t}A = I$ (similarly for $B$): $$ A^{t}B-B^{t}A=A^{t}B-B^{t}A + A^{t}A - B^{t}B=A^{t}(A+B)+B^{t}(A - B) $$

I don't understand how I should continue, and don't see an alternative method.

Answer 1

It results that

\begin{align} A^{t}B-B^{t}A&=A^{t}B-B^{t}A+I-I=\\&=\color{blue}{A^{t}B}\color{red}{-B^{t}A}+\color{blue}{A^{t}A}\color{red}{-B^{t}B}=\\&=\color{blue}{A^{t}(A+B)}\color{red}{-B^{t}(A+B)}=\\&=(A+B)(A^t-B^t)=(A+B)(A-B)^t\,. \end{align}

Since $\;\det(A-B)^t=\det(A-B)\;,\;$ we get that

$$\det(A^{t}B-B^{t}A)=\det(A+B)\det(A-B)\,.$$