Simply the following expressions: a) \( \tan \left(360^{\circ}-x\right) \cdot \sin \left(90^{\circ}-x\right) \) b) \( \frac{\sin \left(180^{\circ}-\alpha\right) \cdot \cos \left(180^{\circ}+\alpha\right)}{\cos \left(90^{\circ}-\alpha\right)} \) c) \( \frac{\cos \left(90^{\circ}+x\right)+\cos \left(180^{\circ}-x\right)+\sin \left(180^{\circ}-x\right)}{\cos \left(360^{\circ}-x\right)} \)

Let's simplify each of the given expressions one at a time. a) \[ \tan \left(360^{\circ}-x\right) \cdot \sin \left(90^{\circ}-x\right) \] Using the periodic properties of the tangent function, we have: \[ \tan \left(360^{\circ}-x\right) = \tan(-x) = -\tan x \] And for the sine function: \[ \sin \left(90^{\circ}-x\right) = \cos x \] Putting it all together: \[ \tan \left(360^{\circ}-x\right) \cdot \sin \left(90^{\circ}-x\right) = -\tan x \cdot \cos x = -\frac{\sin x}{\cos x} \cdot \cos x = -\sin x \] b) \[ \frac{\sin \left(180^{\circ}-\alpha\right) \cdot \cos \left(180^{\circ}+\alpha\right)}{\cos \left(90^{\circ}-\alpha\right)} \] Using sin and cos transformations: \[ \sin \left(180^{\circ}-\alpha\right) = \sin \alpha \] \[ \cos \left(180^{\circ}+\alpha\right) = -\cos \alpha \] \[ \cos \left(90^{\circ}-\alpha\right) = \sin \alpha \] Thus, the expression simplifies to: \[ \frac{\sin \alpha \cdot (-\cos \alpha)}{\sin \alpha} = -\cos \alpha \quad (\text{for } \sin \alpha \neq 0) \] c) \[ \frac{\cos \left(90^{\circ}+x\right)+\cos \left(180^{\circ}-x\right)+\sin \left(180^{\circ}-x\right)}{\cos \left(360^{\circ}-x\right)} \] Calculating each part: \[ \cos(90^{\circ}+x) = -\sin x,\quad \cos(180^{\circ}-x) = -\cos x,\quad \sin(180^{\circ}-x) = \sin x \] So the numerator becomes: \[ -\sin x - \cos x + \sin x = -\cos x \] For the denominator, \[ \cos(360^{\circ}-x) = \cos x \] Thus, we get: \[ \frac{-\cos x}{\cos x} = -1 \quad (\text{for } \cos x \neq 0) \] Final results: a) \(-\sin x\) b) \(-\cos \alpha\) c) \(-1\)