(1) \( \frac{\cos (180+x) \tan (180-x)}{\sin (180-x)} \)

Alright! Let's dive into this trigonometric expression. First, we can simplify \( \cos(180 + x) \) using the cosine addition formula: \[ \cos(180 + x) = -\cos(x) \] Then, we simplify \( \tan(180 - x) \): \[ \tan(180 - x) = -\tan(x) \] And for \( \sin(180 - x) \), it simplifies to: \[ \sin(180 - x) = \sin(x) \] Now plugging these back into the original expression: \[ \frac{\cos(180+x) \tan(180-x)}{\sin(180-x)} = \frac{(-\cos(x))(-\tan(x))}{\sin(x)} = \frac{\cos(x) \tan(x)}{\sin(x)} \] Since \( \tan(x) = \frac{\sin(x)}{\cos(x)} \), we have: \[ \frac{\cos(x) \cdot \frac{\sin(x)}{\cos(x)}}{\sin(x)} \] This simplifies to: \[ \frac{\sin(x)}{\sin(x)} = 1 \] So, the final result is: \[ \boxed{1} \]