Prove that the determinant $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&{\sin \theta }&{\cos \theta }\\{ - \sin \theta }&{ - x}&1\\{\cos \theta }&1&x\end{array}} \right|$ is independent of $\theta$.

Let $\Delta = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&{\sin \theta }&{\cos \theta }\\{ - \sin \theta }&{ - x}&1\\{\cos \theta }&1&x\end{array}} \right|$

Expanding by ${R_1},$

we get
$= x\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - x}&1\\1&x\end{array}} \right| - \sin \theta \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - \sin \theta }&1\\{\cos \theta }&x\end{array}} \right| + \cos \theta \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - \sin \theta }&{ - x}\\{\cos \theta }&1\end{array}} \right|$

$= x( - {x^2} - 1) - \sin \theta ( - x\sin \theta - \cos \theta ) + \cos \theta ( - \sin \theta + x\cos \theta )$

$= - {x^3} - x + x{\sin ^2}\theta + \sin \theta \cos \theta - \sin \theta \cos \theta + x{\cos ^2}\theta$

$= - {x^3} - x + x({\sin ^2}\theta + {\cos ^2}\theta ) = - {x^3} - x + x(1) = - {x^3}$

which is independent of $\theta .$