Simplify the following: (a) \( \frac{\cos \left(180^{\circ}-\theta\right) \cdot \cos \left(90^{\circ}-\theta\right)}{\sin \left(90^{\circ}+\theta\right) \cdot \sin \left(-\theta-180^{\circ}\right)} \)

To simplify the expression \[ \frac{\cos \left(180^{\circ}-\theta\right) \cdot \cos \left(90^{\circ}-\theta\right)}{\sin \left(90^{\circ}+\theta\right) \cdot \sin \left(-\theta-180^{\circ}\right)}, \] we can use trigonometric identities. 1. Start with the numerator: - \(\cos(180^{\circ}-\theta) = -\cos(\theta)\) (since cosine is negative in the second quadrant). - \(\cos(90^{\circ}-\theta) = \sin(\theta)\) (co-function identity). Thus, the numerator becomes: \[ \cos(180^{\circ}-\theta) \cdot \cos(90^{\circ}-\theta) = -\cos(\theta) \cdot \sin(\theta). \] 2. Now for the denominator: - \(\sin(90^{\circ}+\theta) = \cos(\theta)\) (co-function identity). - \(\sin(-\theta-180^{\circ}) = -\sin(-\theta) = \sin(\theta)\) (since sine is odd and \(\sin(\theta + 180^{\circ}) = -\sin(\theta)\)). Therefore, the denominator simplifies to: \[ \sin(90^{\circ}+\theta) \cdot \sin(-\theta-180^{\circ}) = \cos(\theta) \cdot \sin(\theta). \] 3. Now we can substitute everything back into the original expression: \[ \frac{-\cos(\theta) \cdot \sin(\theta)}{\cos(\theta) \cdot \sin(\theta)}. \] 4. Since \(\cos(\theta) \cdot \sin(\theta)\) is common in both the numerator and denominator (assuming \(\theta\) does not make either of them zero), this simplifies to: \[ -1. \] Therefore, the simplified expression is: \[ \boxed{-1}. \]