Using the parametrization $$ A= \begin{pmatrix} a^2+b^2-c^2-d^2&2bc-2ad &2bd+2ac \cr 2bc+2ad &a^2-b^2+c^2-d^2&2cd-2ab \cr 2bd-2ac &2cd+2ab &a^2-b^2-c^2+d^2\\ \end{pmatrix}, $$

we have $$ (1-tr (A))^2-\frac{1}{2} tr ((A-A^t)^2)-4= $$

$$ (9a^2 + b^2 + c^2 + d^2 + 3)(a^2 + b^2 + c^2 + d^2 - 1)=0, $$ since $a^2+b^2+c^2+d^2=1$.

Using the parametrization $$ A= \begin{pmatrix} a^2+b^2-c^2-d^2&2bc-2ad &2bd+2ac \cr 2bc+2ad &a^2-b^2+c^2-d^2&2cd-2ab \cr 2bd-2ac &2cd+2ab &a^2-b^2-c^2+d^2\\ \end{pmatrix}, $$

which comes from quaternions of unit length we have $$ (1-tr (A))^2-\frac{1}{2} tr ((A-A^t)^2)-4= $$

$$ (9a^2 + b^2 + c^2 + d^2 + 3)(a^2 + b^2 + c^2 + d^2 - 1)=0, $$ since $a^2+b^2+c^2+d^2=1$.