In Cartesian coordinates on the sphere and on the plane, the projection and its inverse are given by the formulas
It is not hard to see that
Now we use the trick introduced in this entry. Let
The determinant we are trying to compute is
which is the characteristic polynomial of evaluated at times .
Now, is certainly diagonalizable (which doesn’t even matter, but it makes it easier to think about), and we know its eigenvalues. Why do we know its eigenvalues? Because is a matrix of rank , hence nullity , hence of its eigenvalues are zero. What is the other eigenvalue? It’s the same as the sum of the eigenvalues, which is the trace of , which is . Put that information together, and we have that the characteristic polynomial of is
Substitute for to get