$\begin{aligned} & \begin{aligned}|A| & =\left|\begin{array}{cc}1 & \tan x \\ -\tan x & 1\end{array}\right| \\ & =1+\tan ^2 x \neq 0\end{aligned} \\ & \text { If } A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \text { and ad }-b c \neq 0, \\ & \text { then } A^{-1}=\frac{1}{(a d-b c)}\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]\end{aligned}$
$\begin{aligned} & \therefore \quad \mathrm{A}^{-1}=\frac{1}{1+\tan ^2 x}\left[\begin{array}{cc}1 & -\tan x \\ \tan x & 1\end{array}\right] \\ & \therefore \quad \mathrm{A}^{\mathrm{T}} \mathrm{A}^{-1}=\left[\begin{array}{cc}1 & -\tan x \\ \tan x & 1\end{array}\right]\left[\begin{array}{cc}\frac{1}{1+\tan ^2 x} & \frac{-\tan x}{1+\tan ^2 x} \\ \frac{\tan x}{1+\tan ^2 x} & \frac{1}{1+\tan ^2 x}\end{array}\right]\end{aligned}$
$\begin{aligned} & =\left[\begin{array}{ll}\frac{1-\tan ^2 x}{1+\tan ^2 x} & \frac{-2 \tan x}{1+\tan ^2 x} \\ \frac{2 \tan x}{1+\tan ^2 x} & \frac{1-\tan ^2 x}{1+\tan ^2 x}\end{array}\right] \\ & =\left[\begin{array}{ll}\cos 2 x & -\sin 2 x \\ \sin 2 x & \cos 2 x\end{array}\right]\end{aligned}$