\( \begin{array}{ll}5.2 & \text { Simplify the following without using a calculator: } \\ & \frac{\sin \left(90^{\circ}-\theta\right) \cdot \cos 480^{\circ}+\cos \left(180^{\circ}-\theta\right)}{\cos \theta \cdot \sin 150^{\circ}-\tan 180^{\circ}} \\ 5.3 & \text { Prove that } \frac{\cos x}{\sin 2 x}-\frac{\cos 2 x}{2 \sin x}=\sin x\end{array} \)

To simplify the expression given, first, let's break down the components. The sine and cosine identities tell us that \( \sin(90^\circ - \theta) = \cos \theta \), and \( \cos(480^\circ) = \cos(480^\circ - 360^\circ) = \cos(120^\circ) = -\frac{1}{2} \). Thus, we can rewrite the expression as: \[ \frac{\cos \theta \cdot (-\frac{1}{2}) + \cos(180^\circ - \theta)}{\cos \theta \cdot \frac{1}{2} - 0} \] Knowing that \( \cos(180^\circ - \theta) = -\cos \theta \), the numerator simplifies to: \[ -\frac{1}{2} \cos \theta - \cos \theta = -\frac{3}{2} \cos \theta \] The denominator simplifies to \( \frac{1}{2} \cos \theta \). Hence, the whole expression simplifies to: \[ \frac{-\frac{3}{2} \cos \theta}{\frac{1}{2} \cos \theta} = -3 \] Now onto the proof for the second part. We start with the left side \[ \frac{\cos x}{\sin 2x} - \frac{\cos 2x}{2 \sin x} \] Using the double angle identity, \( \sin 2x = 2 \sin x \cos x \), we rewrite the first term: \[ \frac{\cos x}{2 \sin x \cos x} = \frac{1}{2 \sin x} \] The second term can be rewritten as: \[ \frac{\cos 2x}{2 \sin x} \] Now, let's focus on combining them: \[ \frac{1 - \cos 2x}{2 \sin x} \] Using the identity \( 1 - \cos 2x = 2 \sin^2 x \), we substitute: \[ \frac{2 \sin^2 x}{2 \sin x} = \sin x \] Thus, proving that \[ \frac{\cos x}{\sin 2x} - \frac{\cos 2x}{2 \sin x} = \sin x \] And that completes the proof!