The Characteristic Equation will be:
$\begin{vmatrix} \cos \theta - \lambda & \sin \theta\\ \sin \theta & -\cos \theta -\lambda \end{vmatrix} = 0\\ \\ \Rightarrow \lambda^{2} - \cos^{2}\theta - \sin^{2}\theta = 0 \\ \Rightarrow \lambda^{2} - 1 = 0 \\ \Rightarrow \lambda = \pm 1$
Here, the $\lambda$'s are the eigen values of the given matrix. Let us now calculate the eigen vectors. FYI, a square matrix of order $n$ is diagonalizable iff it has $n$ linearly independent eigen vectors. Also, the union of the zero vector and the set of all eigenvectors corresponding to eigenvalues $\lambda$ is known as the eigenspace of the matrix.
For $\lambda = 1$,
$\begin{bmatrix} \cos \theta - 1 & \sin \theta\\ \sin \theta & -\cos \theta - 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix} $
Here, $X = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ is the eigen vector and let $A = \begin{bmatrix} \cos \theta - 1 & \sin \theta\\ \sin \theta & -\cos \theta - 1 \end{bmatrix}$
Performing the row operation $R_1 \rightarrow \cos \theta \times R_1 + \sin \theta \times R_2$ we get
$\begin{bmatrix}1 - \cos \theta & -\sin \theta\\ \sin \theta & -\cos \theta - 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$
Let us call the coefficient matrix above $A'$.
Now, since $AX = 0$ & $A'X=0$ $\Rightarrow (A + A')X = 0$
$\Rightarrow \begin{bmatrix}0 & 0 \\ \sin \theta & -\cos \theta - 1 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$
From this, we have a relation that $ (\sin \theta) x_1 + (-\cos \theta - 1)x_2 = 0$
Let us assume $x_2 = c_1$, So, $x_1 = c_1(1+ \cos\theta)/\sin\theta$
Therefore, our eigen vector corresponding to $\lambda = 1$ is of the form $X_1 = c_1\begin{bmatrix} (1+ \cos\theta)/\sin\theta\\ 1 \end{bmatrix}$
In the same way, for $\lambda = -1$, we have
$\begin{bmatrix} \cos \theta +1 & \sin \theta\\ \sin \theta & -\cos \theta + 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix} $
Performing the row operation $R_2 \rightarrow \cos \theta \times R_1 + \sin \theta \times R_2$ on $BX = 0$ and then $R_2 \rightarrow R_2 - R_1$ we get
$\begin{bmatrix}1 + \cos \theta & \sin \theta\\ 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$
From this, we get the relation $ (1 + \cos\theta)x_1 + (\sin\theta)x_2 = 0 $. Let us assume $x_2 = c_2$, so we get $x_1 = -c_2\sin\theta/(1 + \cos\theta) $.
Therefore, our eigen vector corresponding to $\lambda = -1$ is of the form $X_2 = c_2\begin{bmatrix} -\sin\theta/(1 + \cos\theta)\\ 1 \end{bmatrix}$
The matrix A will be diagonalizable iff $X_1$ and $X_2$ are linearly independent. Let us check if there arises a case where $X_1$ and $X_2$ are linearly dependent.
$\Rightarrow c_1\begin{bmatrix} (1+ \cos\theta)/\sin\theta\\ 1 \end{bmatrix} + c_2\begin{bmatrix} -\sin\theta/(1 + \cos\theta)\\ 1 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$
From, this we get that for linear dependence
$c_1 + c_2 = 0 $ (equation 1) and,
$ c_1(1+ \cos\theta)/\sin\theta - c_2\sin\theta/(1 + \cos\theta) = 0 $ (equation 2)
such that $c_1, c_2 \neq 0$.
Substituting the relation between $c_1 $& $ c_2$ from equation 1 into equation 2, we get the relation,
$ (1+ \cos\theta)/\sin\theta = -\sin\theta/(1 + \cos\theta) $
$ \Rightarrow (1+ \cos\theta)^{2} = -\sin\theta^{2} $
$ \Rightarrow 1+ 2\cos\theta + \cos^{2}\theta = \cos^{2}\theta - 1 $
$ \Rightarrow (1+ \cos\theta) = 0 $
$ \Rightarrow \cos\theta = -1$.
$\therefore$ when $\theta = (2n+1)\pi$, where $n$ is an integer, $A$ will not be diagonalizable. Otherwise, $A$ will be diagonalizable. The eigenspace of A will be the set spanned by $X_1$ and $X_2$.
An interesting fact: As pointed out by @ankitgupta.1729, since the eigen values are different, the matrix will always be diagonalizable. But, the point where $\cos\theta + 1 = 0$, the eigen vectors become undefined. This is the point where the matrix $A$ becomes non-diagonalisable.