What is the determinant of the rotation matrix R_theta = & - & ?

The matrix $R_\theta$ is given as:

$$R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$

For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is calculated as $ad - bc$.
Here, $a = \cos\theta$, $b = -\sin\theta$, $c = \sin\theta$, $d = \cos\theta$.
Applying this formula to matrix $R_\theta$:

$$\det(R_\theta) = (\cos\theta)(\cos\theta) - (-\sin\theta)(\sin\theta)$$ $$\det(R_\theta) = \cos^2\theta - (-\sin^2\theta)$$ $$\det(R_\theta) = \cos^2\theta + \sin^2\theta$$

The fundamental trigonometric identity states that $\sin^2\theta + \cos^2\theta = 1$.
Therefore:

$$\det(R_\theta) = 1$$

The calculated determinant is 1. This matches option D.

The final answer is $\text{D. 1}$.