Find all possible values of the determinant of
\[\begin{pmatrix} \sec^2 x & 1 & 1 \\ \cos^2 x & \cos^2 x & \csc^2 x \\ 1 & \cos^2 x & \cot^2 x \end{pmatrix},\]as $x$ ranges over all real numbers (where the determinant is defined).
Expanding the determinant, we obtain
\begin{align*}
\begin{vmatrix} \sec^2 x & 1 & 1 \\ \cos^2 x & \cos^2 x & \csc^2 x \\ 1 & \cos^2 x & \cot^2 x \end{vmatrix} &= \sec^2 x \begin{vmatrix} \cos^2 x & \csc^2 x \\ \cos^2 x & \cot^2 x \end{vmatrix} - \begin{vmatrix} \cos^2 x & \csc^2 x \\ 1 & \cot^2 x \end{vmatrix} + \begin{vmatrix} \cos^2 x & \cos^2 x \\ 1 & \cos^2 x \end{vmatrix} \\
&= \sec^2 x (\cos^2 x \cot^2 x - \csc^2 x \cos^2 x) - (\cos^2 x \cot^2 x - \csc^2 x) + (\cos^4 x - \cos^2 x) \\
&= \frac{1}{\cos^2 x} \left( \cos^2 x \cdot \frac{\cos^2 x}{\sin^2 x} - \frac{1}{\sin^2 x} \cdot \cos^2 x \right) - \left( \cos^2 x \cdot \frac{\cos^2 x}{\sin^2 x} - \frac{1}{\sin^2 x} \right) + (\cos^4 x - \cos^2 x) \\
&= \frac{\cos^2 x - 1}{\sin^2 x} - \frac{\cos^2 x}{\sin^2 x} (\cos^2 x - 1) + \cos^4 x - \cos^2 x \\
&= \frac{-\sin^2 x}{\sin^2 x} - \frac{\cos^2 x}{\sin^2 x} (-\sin^2 x) + \cos^4 x - \cos^2 x \\
&= -1 + \cos^2 x + \cos^4 x - \cos^2 x \\
&= \cos^4 x.
\end{align*}The range of $\cos^4 x$ is $[0,1].$  However, if $\cos^4 x = 0,$ then $\cos x = 0,$ which means $\sec x$ is not defined.  And if $\cos^4 x = 1,$ then $\cos^2 x =1,$ so $\sin^2 x = 0,$ which means $\csc x$ is not defined.  Therefore, the set of all possible values of the determinant is $\boxed{(0,1)}.$