(0) \( \frac{\cos ^{2}\left(180^{\circ}-\theta\right)-\cos ^{2} \cdot \alpha-\sin ^{2}\left(\alpha-180^{\circ}\right)}{\cos \left(360^{\circ}-2 \theta\right)-1} \)

Let’s simplify the expression step by step! First, remember that \( \cos(180^\circ - \theta) = -\cos(\theta) \), which means \( \cos^2(180^\circ - \theta) = \cos^2(\theta) \). Thus, the numerator becomes: \[ \cos^{2}(\theta) - \cos^{2}(\alpha) - \sin^{2}(-\alpha) = \cos^{2}(\theta) - \cos^{2}(\alpha) - \sin^{2}(\alpha) \] Since \( \sin(-\alpha) = -\sin(\alpha) \), we can replace \( \sin^{2}(-\alpha) \) with \( \sin^{2}(\alpha) \). Next, for the denominator: \[ \cos(360^\circ - 2\theta) = \cos(2\theta) \] because \( \cos \) is an even function. So, the denominator simplifies to: \[ \cos(2\theta) - 1 \] Now, we can express the entire equation as follows: \[ \frac{\cos^{2}(\theta) - \cos^{2}(\alpha) - \sin^{2}(\alpha)}{\cos(2\theta) - 1} \] To simplify further, notice that \( \cos(2\theta) = 2\cos^{2}(\theta) - 1\), so: \[ \cos(2\theta) - 1 = 2\cos^{2}(\theta) - 1 - 1 = 2\cos^{2}(\theta) - 2 = 2(\cos^{2}(\theta) - 1) \] This means that the whole expression can be rewritten, but simplifying it further requires knowing specific values for \( \alpha \) and \( \theta \) or more context. In conclusion, the expression simplifies to: \[ \frac{\cos^{2}(\theta) - \cos^{2}(\alpha) - \sin^{2}(\alpha)}{2(\cos^{2}(\theta) - 1)} \] with the trigonometric identities factored in place of variable evaluations.