Explanation

(b)
$|A| = \begin{bmatrix}
1 & \tan x \\
- \tan x & 1
\end{bmatrix} = 1 + \tan^{2}x \neq 0$
So A is invertible. Also,
adj $A = \begin{bmatrix}
1 & \tan x \\
- \tan x & 1
\end{bmatrix}^{T} = \begin{bmatrix}
1 & - \tan x \\
\tan x & 1
\end{bmatrix}$
Now,
$A^{- 1} = \frac{1}{|A|}adj\ A$
$\Rightarrow A^{- 1} = \frac{1}{1 + \tan^{2}x}\begin{bmatrix}
1 & - \tan x \\
\tan x & 1
\end{bmatrix}$
$= \begin{bmatrix}
\frac{1}{1 + \tan^{2}x} & \frac{{- tan}x}{1 + \tan^{2}x} \\
\frac{\tan x}{1 + \tan^{2}x} & \frac{1}{1 + \tan^{2}x}
\end{bmatrix}$
$\therefore A^{T}A^{- 1} = \begin{bmatrix}
1 & - \tan x \\
\tan x & 1
\end{bmatrix}\begin{bmatrix}
\frac{1}{1 + \tan^{2}x} & \frac{{- tan}x}{1 + \tan^{2}x} \\
\frac{\tan x}{1 + \tan^{2}x} & \frac{1}{1 + \tan^{2}x}
\end{bmatrix}$
$= \begin{bmatrix}
\frac{1 - \tan^{2}x}{1 + \tan^{2}x} & \frac{{- 2tan}x}{1 + \tan^{2}x} \\
\frac{{2tan}x}{1 + \tan^{2}x} & \frac{1 - \tan^{2}x}{1 + \tan^{2}x}
\end{bmatrix}$
$= \begin{bmatrix}
\cos{2x} & - \sin{2x} \\
\sin{2x} & \cos{2x}
\end{bmatrix}$