Prove the identity \( \frac{\sin ^{2}(180+\theta)+\cos (-\theta) \cdot \sin (90-\theta)}{\sin \theta}+\frac{1}{\tan \theta}=\frac{1+\cos \theta}{\sin \theta} \)

To prove the identity \[ \frac{\sin ^{2}(180+\theta)+\cos (-\theta) \cdot \sin (90-\theta)}{\sin \theta}+\frac{1}{\tan \theta}=\frac{1+\cos \theta}{\sin \theta}, \] we start by simplifying the left-hand side. We know the following trigonometric identities: 1. \(\sin(180 + \theta) = -\sin \theta\). 2. \(\cos(-\theta) = \cos \theta\). 3. \(\sin(90 - \theta) = \cos \theta\). 4. \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), thus \(\frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}\). Using these, we can simplify the left-hand side: \[ \sin^2(180 + \theta) = \sin^2(-\theta) = \sin^2 \theta, \] so the first term becomes: \[ \frac{\sin^2 \theta + \cos \theta \cdot \cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta} = \frac{1}{\sin \theta} \] (the Pythagorean identity states \(\sin^2 \theta + \cos^2 \theta = 1\)). Now we add the second term \(\frac{1}{\tan \theta}\): \[ \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}. \] So now the left-hand side becomes: \[ \frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta} = \frac{1 + \cos \theta}{\sin \theta}. \] Now, the left-hand side is \[ \frac{1 + \cos \theta}{\sin \theta}. \] As we can see, both sides of the original identity are equal, thus we have: \[ \frac{1 + \cos \theta}{\sin \theta} = \frac{1 + \cos \theta}{\sin \theta}. \] The identity is proven.