Answer
Yes, the matrix $A$ is invertible
and its inverse is $A^{-1}=\left[\begin{array}{cc}
sec\theta & -tan \theta\\
-tan \theta& sec \theta
\end{array}\right]$
Work Step by Step
Given
$A=\left[\begin{array}{cc}
sec\theta & tan \theta\\
tan \theta& sec \theta
\end{array}\right]$
then
$|A|=(sec\theta)^{2}-(tan\theta)^{2}=1\ne0 $
thus the matrix $A$ is nonsingular and invertible
therefore $A^{-1}=\frac{1}{|A|} *\left[\begin{array}{cc}
sec\theta & -tan \theta\\
-tan \theta& sec \theta
\end{array}\right]=\left[\begin{array}{cc}
sec\theta & -tan \theta\\
-tan \theta& sec \theta
\end{array}\right]$
