Simplify the following: (a) \( \frac{\cos \left(180^{\circ}-\theta\right) \cdot \cos \left(90^{\circ}-\theta\right)}{\sin \left(90^{\circ}+\theta\right) \cdot \sin \left(-\theta-180^{\circ}\right)} \)
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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 apply the trigonometric identities step by step. 1. **Using angle identities:** - For \(\cos(180^{\circ} - \theta)\), we have: \[ \cos(180^{\circ} - \theta) = -\cos(\theta). \] - For \(\cos(90^{\circ} - \theta)\): \[ \cos(90^{\circ} - \theta) = \sin(\theta). \] - For \(\sin(90^{\circ} + \theta)\): \[ \sin(90^{\circ} + \theta) = \cos(\theta). \] - For \(\sin(-\theta - 180^{\circ})\): \[ \sin(-\theta - 180^{\circ}) = -\sin(-\theta) = \sin(\theta). \] 2. **Substituting these into the original expression:** Now we substitute these identities into our expression: \[ \frac{-\cos(\theta) \cdot \sin(\theta)}{\cos(\theta) \cdot \sin(\theta)}. \] 3. **Simplifying the fraction:** If we assume that \(\cos(\theta) \cdot \sin(\theta) \neq 0\), we can cancel \(\cos(\theta) \cdot \sin(\theta)\) from the numerator and the denominator: \[ = \frac{-1}{1} = -1. \] Thus, the simplified expression is \[ \boxed{-1}. \]