2. \( \frac{\sin \theta-\tan \theta \cdot \cos ^{2} \theta}{\cos \theta-1+\sin ^{2} \theta}=\tan \theta \)

To solve \( \frac{\sin \theta-\tan \theta \cdot \cos ^{2} \theta}{\cos \theta-1+\sin ^{2} \theta}=\tan \theta \), let's start by simplifying the equation step by step. First, recall that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Substituting for \( \tan \theta \): \[ \frac{\sin \theta - \frac{\sin \theta}{\cos \theta} \cdot \cos^2 \theta}{\cos \theta - 1 + \sin^2 \theta} = \frac{\sin \theta}{\cos \theta} \] This simplifies to: \[ \frac{\sin \theta - \sin \theta \cdot \cos \theta}{\cos \theta - 1 + \sin^2 \theta} = \frac{\sin \theta}{\cos \theta} \] Now, we can factor out \( \sin \theta \) from the numerator: \[ \frac{\sin \theta (1 - \cos \theta)}{\cos \theta - 1 + \sin^2 \theta} = \frac{\sin \theta}{\cos \theta} \] At this point, notice that \( \cos \theta - 1 + \sin^2 \theta \) can be rewritten using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \): \[ \cos \theta - 1 + (1 - \cos^2 \theta) = -\cos \theta + 1 \] Thus we have: \[ \frac{\sin \theta (1 - \cos \theta)}{-\cos \theta + 1} = \frac{\sin \theta}{\cos \theta} \] Which simplifies to: \[ \sin \theta = \sin \theta * \frac{\sin \theta}{\cos \theta} \] By cross-multiplying and ensuring \( \sin \theta \neq 0 \): \[ \sin \theta \cdot \cos \theta = \sin^2 \theta \] Dividing by \( \sin \theta \) (assuming \( \sin \theta \neq 0 \)) gives us: \[ \cos \theta = \sin \theta \] This implies: \[ \tan \theta = 1 \] The solutions occur when \( \theta = \frac{\pi}{4} + k\pi \) for any integer \( k \). Lastly, ensure the \( \sin \theta = 0 \) case stands out as well, which leads to: \[ \theta = n\pi \text{ for any integer } n. \] Thus, the complete set of solutions comprises: \[ \theta = \frac{\pi}{4} + k\pi \quad \text{and} \quad \theta = n\pi \] for any integers \( k \) and \( n \).