Since given matrix $A$ is invertible $\Rightarrow \left|A\right|\ne 0$
 

$\left|A\right|=e^{-t}\left|\begin{array}{ccc}1& \cos t& \sin t\\ 1& -\cos t-\sin t& -\sin t+\cos t\\ 1& 2 \sin t& -2 \cos t\end{array}\right|$

$R_2\to R_2-R_1$

$R_3\to R_3-R_1$

$=e^{-t}\left|\begin{array}{ccc}1& \cos t& \sin t\\ 0& -2 \cos t-\sin t& -2 \sin t+\cos t\\ 0& 2 \sin t-\cos t& -2 \cos t-\sin t\end{array}\right|$

$=e^{-t}\left\{{\left(2\cos t+\sin t\right)}^2+{\left(2\sin t-\cos t\right)}^2\right\}$

$=5e^{-t}$

$\left|A\right|=5 e^{-t}\ne 0 \forall \mathrm{ }\mathrm{t}\in \mathrm{R}$

Hence, given matrix is always invertible.