How prove that $\;(1-\mathrm{Tr}\,A)^2+\sum_{1\le i\le j\le 3}(a_{ij}-a_{ji})^2=4\;\;?$ Let $A=\begin{bmatrix}
a_{11}&a_{12}&a_{13}\\
a_{21}&a_{22}&a_{23}\\
a_{31}&a_{32}&a_{33}
\end{bmatrix}$ be an orthogonal matrix with $a_{i,j}\in \mathbb R$, where $\det(A)=1$
Show that
$$(1-\mathrm{Tr}\,A)^2+\sum_{1\le i\le j\le 3}(a_{ij}-a_{ji})^2=4$$
 A: Note that
$$\mathrm{Tr}\left[(A-A^T)^2\right]=-2\sum_{1\le i\le j\le 3}(a_{ij}-a_{ji})^2.$$
Therefore the identity can be rewritten as
$$2(1-\mathrm{Tr}\,A)^2-\mathrm{Tr}\left[(A-A^T)^2\right]=8.$$
Since $\mathrm{Tr}\,AA^T=\mathrm{Tr}\,1=3$, this is equivalent to
$$\left(\mathrm{Tr}\,A\right)^2-\mathrm{Tr}\,A^2-2\mathrm{Tr}\,A=0.$$
The left side of the last relation can be rewritten in terms of eigenvalues $\lambda_{1,2,3}$ of $A$:
$$(\lambda_1+\lambda_2+\lambda_3)^2-(\lambda_2^2+\lambda_2^2+\lambda_3^2)-2(\lambda_1+\lambda_2+\lambda_3).$$
Using that $\lambda_1\lambda_2\lambda_3=\mathrm{det}\,A=1$, we reduce the problem to showing that
$$\lambda_1^{-1}+\lambda_2^{-1}+\lambda_3^{-1}=\lambda_1+\lambda_2+\lambda_3$$
But this follows from $\mathrm{Tr}\,A^{-1}=\mathrm{Tr}\,A^T=\mathrm{Tr}\,A$. $\blacksquare$

In case someone would like to have an "explanation" of the very first identity. The matrix $A-A^T$ can be seen as an element $a$ of Lie algebra $\mathfrak{so}(3)$, which can in turn be seen as a three-dimensional vector. The squared length of this vector coincides with Cartan-Killing norm of $a$ up to a numerical factor. 
