1. **State the problem:** We want to find the matrix $A = BB^T + CC^T$ where $$B = \begin{bmatrix} \cos \theta \\ \sin \theta \end{bmatrix}, \quad C = \begin{bmatrix} \sin \theta \\ -\cos \theta \end{bmatrix}$$ for any real number $\theta$. 2. **Recall the formula:** For vectors $x$ and $y$, the matrix $xx^T$ is the outer product, resulting in a matrix. 3. **Calculate $BB^T$:** $$BB^T = \begin{bmatrix} \cos \theta \\ \sin \theta \end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \end{bmatrix} = \begin{bmatrix} \cos^2 \theta & \cos \theta \sin \theta \\ \sin \theta \cos \theta & \sin^2 \theta \end{bmatrix}$$ 4. **Calculate $CC^T$:** $$CC^T = \begin{bmatrix} \sin \theta \\ -\cos \theta \end{bmatrix} \begin{bmatrix} \sin \theta & -\cos \theta \end{bmatrix} = \begin{bmatrix} \sin^2 \theta & -\sin \theta \cos \theta \\ -\cos \theta \sin \theta & \cos^2 \theta \end{bmatrix}$$ 5. **Add $BB^T$ and $CC^T$:** $$A = BB^T + CC^T = \begin{bmatrix} \cos^2 \theta + \sin^2 \theta & \cos \theta \sin \theta - \sin \theta \cos \theta \\ \sin \theta \cos \theta - \cos \theta \sin \theta & \sin^2 \theta + \cos^2 \theta \end{bmatrix}$$ 6. **Simplify using trigonometric identities:** Recall that $\cos^2 \theta + \sin^2 \theta = 1$ and the off-diagonal terms cancel out: $$A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I_2$$ 7. **Conclusion:** The matrix $A$ is the $2 \times 2$ identity matrix for all real $\theta$. **Final answer:** $$A = I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$