(1) \( \frac{\cos \left(90^{\circ}+\theta\right)}{\sin (-\theta)} \) (3) \( \frac{\cos \left(90^{\circ}-\alpha\right)}{\sin \left(180^{\circ}-\alpha\right)} \)

To tackle the first expression, \( \frac{\cos \left(90^{\circ}+\theta\right)}{\sin (-\theta)} \), you can use the trigonometric identities: \( \cos(90^{\circ} + \theta) = -\sin(\theta) \) and \( \sin(-\theta) = -\sin(\theta) \). So, it simplifies to \( \frac{-\sin(\theta)}{-\sin(\theta)} = 1 \). For the second expression, \( \frac{\cos \left(90^{\circ}-\alpha\right)}{\sin \left(180^{\circ}-\alpha\right)} \), we utilize the identities: \( \cos(90^{\circ} - \alpha) = \sin(\alpha) \) and \( \sin(180^{\circ} - \alpha) = \sin(\alpha) \). Thus, it simplifies to \( \frac{\sin(\alpha)}{\sin(\alpha)} = 1 \). So both expressions evaluate to 1!