$\mathrm{A}=\left[\begin{array}{cc}1& \tan \mathrm{x}\\ -\tan \mathrm{x}& 1\end{array}\right]\text{ and }\mathrm{f}\left(\mathrm{x}\right)\text{ is defined as }\mathrm{f}\left(\mathrm{x}\right)=\det \cdot \left({\mathrm{A}}^{\mathrm{T}}{\mathrm{A}}^{-1}\right)\text{ then the value of }$ $\underset{\text{ntimes }}{\underbrace{\mathrm{f}\left(\mathrm{f}\right(\mathrm{f}(\mathrm{f}\dots \dots \dots ..\mathrm{f}(\mathrm{x}\left)\right)\left)\right)}}\text{ is }(\mathrm{n}\ge 2)$ _____.

$\mathrm{A}=\left[\begin{array}{cc}1& \tan \mathrm{x}\\ -\tan \mathrm{x}& 1\end{array}\right]$
$\text{ Hence, }\det .\mathrm{A}={\sec }^2\mathrm{x}$
$\therefore \det {\mathrm{A}}^{\mathrm{T}}={\sec }^2\mathrm{x}$
$\text{ Now }\mathrm{f}\left(\mathrm{x}\right)=\det .\left({\mathrm{A}}^{\mathrm{T}}{\mathrm{A}}^{-1}\right)$
$\begin{array}{l}=\left(\det \cdot {\mathrm{A}}^{\mathrm{T}}\right)\left(\det \cdot {\mathrm{A}}^{-1}\right)\\ =\left(\det \cdot {\mathrm{A}}^{\mathrm{T}}\right)(\det \cdot \mathrm{A}{)}^{-1}\\ =\frac{\det \cdot \left({\mathrm{A}}^{\mathrm{T}}\right)}{\det \cdot \left(\mathrm{A}\right)}=1\end{array}$
Hence, $\mathrm{f}\left(\mathrm{x}\right)=1$.