Answer
See below
Work Step by Step
Given $$ A=\begin{bmatrix}
\sin t & \cos t & 1 \\ \cos t & -\sin t & 0 \\ \sin t & -\cos t & 0
\end{bmatrix}$$
So, we get
$det (A)=\begin{vmatrix}
\sin t & \cos t & 1 \\ \cos t & -\sin t & 0 \\ \sin t & -\cos t & 0
\end{vmatrix}\\
=\sin t(-\sin t).0+\cos t.0.\sin t+1.\cos t.(-\cos t)-\sin t(-\sin t).1-(-\cos t).0.\sin t-0.\cos t.\cos t\\
=-\cos^2 (t)+\sin^2(t)\\
=-\cos(2t)$
