If $A=\left[\begin{array}{cc}1& \tan x\\ -\tan x& 1\end{array}\right]$, then $A^T{A}^{-1}$ is

The correct option is B $\left[\begin{array}{cc}\cos 2x& -\sin 2x\\ \sin 2x& \cos 2x\end{array}\right]$
$|A|=\left[\begin{array}{cc}1& \tan x\\ -\tan x& 1\end{array}\right]=1+{\tan }^{2}x\ne 0$
So, $A$ is invertible. Also,
$adjA={\left[\begin{array}{cc}1& \tan x\\ -\tan x& 1\end{array}\right]}^T=\left[\begin{array}{cc}1& -\tan x\\ \tan x& 1\end{array}\right]$

Now,
$A^{-1}=\frac{1}{|A|}adjA$
$\Rightarrow A^{-1}=\frac{1}{(1+{\tan }^{2}x)}\left[\begin{array}{cc}1& -\tan x\\ \tan x& 1\end{array}\right]$

$=\left[\begin{array}{cc}\frac{1}{1+{\tan }^{2}x}& \frac{-\tan x}{1+{\tan }^{2}x}\\ \frac{\tan x}{1+{\tan }^{2}x}& \frac{1}{1+{\tan }^{2}x}\end{array}\right]$

$\therefore A^T{A}^{-1}=\left[\begin{array}{cc}1& -\tan x\\ \tan x& 1\end{array}\right]\left[\begin{array}{cc}\frac{1}{1+{\tan }^{2}x}& \frac{-\tan x}{1+{\tan }^{2}x}\\ \frac{\tan x}{1+{\tan }^{2}x}& \frac{1}{1+{\tan }^{2}x}\end{array}\right]$

$=\left[\begin{array}{cc}\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}& \frac{-2\tan x}{1+{\tan }^{2}x}\\ \frac{2\tan x}{1+{\tan }^{2}x}& \frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}\end{array}\right]$

$=\left[\begin{array}{cc}\cos 2x& -\sin 2x\\ \sin 2x& \cos 2x\end{array}\right]$