Prove that a real $2 \times 2$ matrix is orthogonal if and only if it is of one of the following forms: Proof Review And Possible Author Error?

Question

Prove that a real $2 \times 2$ matrix is orthogonal if and only if it is of one of the forms

$\begin{bmatrix} a & b \\ -b & a \end{bmatrix}$, $\begin{bmatrix} a & b \\ b & -a \end{bmatrix}$

where $a^2 + b^2 = 1$.

My understanding is that if and only if represents equivalence. So when we have $A$ if and only if $B$ ($A \Longleftrightarrow B$), this is saying $A \implies B$ and $B \implies A$. Therefore, for this problem, we must prove two things:

1. A real $2 \times 2$ orthogonal matrix has form $\begin{bmatrix} a & b \\ -b & a \end{bmatrix}$ or $\begin{bmatrix} a & b \\ b & -a \end{bmatrix}$.

2. The matrices $\begin{bmatrix} a & b \\ -b & a \end{bmatrix}$ and $\begin{bmatrix} a & b \\ b & -a \end{bmatrix}$ are orthogonal matrices.

I found 2. to be easy, but I'm still unsure as to whether I did 1. correctly.

1.

Let $A = \begin{bmatrix} a & c \\ b & d \end{bmatrix}$ be an orthogonal matrix $\forall a, b, c, d \in \mathbb{R}$ and $a^2 + b^2 = 1$.

$\therefore \begin{bmatrix} a & c \\ b & d \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a^2 + c^2 & ab + cd \\ ab + cd & b^2 + d^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $

We now have $a^2 + c^2 = 1, b^2 + d^2 = 1, ab + cd = 0, a^2 + b^2 = 1$.

$\therefore c^2 - b^2 = 0 \implies \pm c = \pm b$ and $a^2 - d^2 = 0 \implies \pm a = \pm d$.

Since we had $A = \begin{bmatrix} a & c \\ b & d \end{bmatrix}$, we have that $b = \pm c$ and $d = \pm a$.

$Q.E.D.$

2.

$\begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} a & -b \\ b & a \end{bmatrix} = \begin{bmatrix} a^2 + b^2 & ab - ab \\ ab - ab & b^2 + a^2 \end{bmatrix} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \begin{bmatrix} a & b \\ -b & a \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\begin{bmatrix} a & b \\ b & -a \end{bmatrix} \begin{bmatrix} a & b \\ b & -a \end{bmatrix} = \begin{bmatrix} a^2 + b^2 & ab - ab \\ ab - ab & b^2 + a^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Since $a^2 + b^2 = 1$.

$Q.E.D.$

I would greatly appreciate it if people could please take the time to review my proof for correctness.

EDIT (New Proof for 1.):

$A = \begin{bmatrix} a & c \\ b & d \end{bmatrix}$ be an orthogonal matrix $\forall a, b, c, d \in \mathbb{R}$

$\therefore \begin{bmatrix} a & c \\ b & d \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a^2 + c^2 & ab + cd \\ ab + cd & b^2 + d^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$a^2 + c^2 = 1$, $b^2 + d^2 = 1$, $ab + cd = 0$.

Let $a = \cos(\theta)$, $c = \sin(\theta)$, $b = \cos(\phi)$, $d = \sin(\phi)$.

$\cos(\theta) \cos(\phi) + \sin(\theta)\sin(\phi) = 0$

$\implies cos(\theta - \phi) = 0$

$\therefore \theta - \phi = \dfrac{(2k + 1)\pi}{2} \ \forall k \in \mathbb{Z}$

$\phi = \theta - \dfrac{(2k + 1)\pi}{2}$

$b = \cos\left( \theta - \dfrac{(2k + 1)\pi}{2} \right)$ and $d = \sin\left( \theta - \dfrac{(2k + 1)\pi}{2} \right)$.

$\therefore b = \cos(\theta)\cos\left( \dfrac{(2k + 1)\pi}{2} \right) + \sin(\theta)\sin\left( \dfrac{(2k + 1)\pi}{2} \right)$

$= \begin{cases} -\sin(\theta) & \forall \ k \ \text{odd} \\ \sin(\theta) & \forall \ k \ \text{even} \\ \end{cases}$

and

$d = \sin(\theta)\cos\left( \dfrac{(2k + 1)\pi}{2} \right) - \cos(\theta)\sin\left( \dfrac{(2k + 1)\pi}{2} \right)$

$= \begin{cases} \cos(\theta) & \forall \ k \ \text{odd} \\ -\cos(\theta) & \forall \ k \ \text{even} \\ \end{cases}$

$\therefore A = \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{bmatrix}$ if $k$ is odd

or

$\therefore A = \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ \sin(\theta) & -\cos(\theta) \end{bmatrix}$ if $k$ is even

$\therefore A = \begin{bmatrix} a & c \\ -c & a \end{bmatrix}$ if $k$ is odd

or

$\therefore A = \begin{bmatrix} a & c \\ c & -a \end{bmatrix}$ if $k$ is even

$Q.E.D.$