Question

$A=\left[\begin{array}{cc}1& \tan x\\ -\tan x& 1\end{array}\right]$ and $f\left(x\right)$ is defined as $f\left(x\right)=det.A^T{A}^{-1}$ then the value of $\frac{f\left(f\right(f......f\left(x\right)\left)\right))}{n\text{times}}(\text{for}n\ge 2)$ is

Answer 1

$A=\left[\begin{array}{cc}1& \tan x\\ -\tan x& 1\end{array}\right]$
Hence, $detA={\sec }^{2}x$
$\therefore det{A}^T={\sec }^{2}x$

Now,
$f\left(x\right)=det\left(A^T{A}^{-1}\right)$
$=(det{A}^T)(det{A}^{-1})$
$=(det{A}^T)(detA{)}^{-1}$
$=\frac{det\left(A^T\right)}{det\left(A\right)}$
$=1$
Hence, $f\left(x\right)=1$