(4) \( \frac{\sin \theta \cos \theta}{1+\cos \theta}=\frac{1-\cos \theta}{\tan \theta} \)
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To solve the equation \( \frac{\sin \theta \cos \theta}{1+\cos \theta}=\frac{1-\cos \theta}{\tan \theta} \), we can clarify it step-by-step: First, recall that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Thus, we can rewrite the right-hand side: \[ \frac{1 - \cos \theta}{\tan \theta} = \frac{1 - \cos \theta}{\frac{\sin \theta}{\cos \theta}} = \frac{(1 - \cos \theta) \cos \theta}{\sin \theta} \] Now the equation becomes: \[ \frac{\sin \theta \cos \theta}{1+\cos \theta} = \frac{(1 - \cos \theta) \cos \theta}{\sin \theta} \] Cross-multiplying gives us: \[ \sin^2 \theta \cos \theta = (1 - \cos \theta)(1 + \cos \theta) = 1 - \cos^2 \theta = \sin^2 \theta \] So we have: \[ \sin^2 \theta \cos \theta = \sin^2 \theta \] Assuming \( \sin^2 \theta \neq 0 \) (otherwise \( \theta = n\pi \) for \( n \in \mathbb{Z} \)), we can safely divide both sides by \( \sin^2 \theta \): \[ \cos \theta = 1 \] This means: \[ \theta = 2n\pi \quad (n \in \mathbb{Z}) \] Thus, the complete solution to the equation is \( \theta = 2n\pi \) for any integer \( n \). So, the solutions are \( \theta = 0, 2\pi, 4\pi, \ldots \) and so on.
