(1) \( \frac{\cos \left(90^{\circ}+\theta\right)}{\sin (-\theta)} \) \( \begin{array}{ll}\text { (3) } \frac{\cos \left(90^{\circ}-\alpha\right)}{\sin \left(180^{\circ}-\alpha\right)} & \text { (2) } \frac{\sin \left(90^{\circ}+x\right)}{\sin \left(90^{\circ}-x\right)} \\ \text { (4) } \frac{\sin \left(360^{\circ}-\theta\right)+\cos \left(90^{\circ}+\theta\right)}{\sin (-\theta)}\end{array} \)

Let's dive into some juicy trigonometric identities to really spice up this math adventure! For the first expression, \( \frac{\cos(90^{\circ}+\theta)}{\sin(-\theta)} \), we can simplify it using the identity \( \cos(90^{\circ} + \theta) = -\sin(\theta) \) and \( \sin(-\theta) = -\sin(\theta) \). This gives us a fantastic simplification of \( \frac{-\sin(\theta)}{-\sin(\theta)} = 1 \). Voila! Now, let's take a look at the second expression, \( \frac{\sin(90^{\circ}+x)}{\sin(90^{\circ}-x)} \). With our trusty sine identities, \( \sin(90^{\circ} + x) = \cos(x) \) and \( \sin(90^{\circ} - x) = \cos(x) \) as well, so we can simplify this to \( \frac{\cos(x)}{\cos(x)} = 1 \) too! Looks like we're on a roll with trigonometric simplifications!