Answer
See explanation
Work Step by Step
Multiply matrices to see final transformation matrix:
\[
\begin{array}{l}
A_{2} A_{1}=\left[\begin{array}{ccc}
\sec \varphi & -\tan \varphi & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \cdot\left[\begin{array}{ccc}
1 & 0 & 0 \\
\sin \varphi & \cos \varphi & 0 \\
0 & 0 & 1
\end{array}\right]= \\
{\left[\begin{array}{cccc}
\sec \varphi-\tan \varphi \sin \varphi & -\tan \varphi \cos \varphi & 0 \\
\sin \varphi & 0 & \cos \varphi & 0 \\
0 & 0 & 1
\end{array}\right]}
\end{array}
\]
Calculate these expressions:
\[
\sec \varphi-\tan \varphi \sin \varphi=\frac{1}{\cos \varphi}-\frac{\sin \varphi}{\cos \varphi} \sin \varphi=\frac{1-\sin ^{2} \varphi}{\cos \varphi}=\frac{\cos ^{2} \varphi}{\cos \varphi}=\cos \varphi
\]
\[
-\tan \varphi \cos \varphi=-\frac{\sin \varphi}{\cos \varphi} \cos \varphi=-\sin \varphi
\]
Meaning that the last matrix can be written as:
\[
\left[\begin{array}{ccc}
\cos \varphi & -\sin \varphi & 0 \\
\sin \varphi & \cos \varphi & 0 \\
0 & 0 & 1
\end{array}\right]
\]
This matrix represents a rotation in $\mathbb{R}^{2}$ in homogeneous coordinates.
