47. \( \frac{\sin \beta}{1-\cos \beta}=\frac{1+\cos \beta}{\sin \beta} \)

To solve the equation \( \frac{\sin \beta}{1-\cos \beta}=\frac{1+\cos \beta}{\sin \beta} \), we can start by cross-multiplying to eliminate the fractions: \[ \sin^2 \beta = (1 - \cos \beta)(1 + \cos \beta) \] The right side simplifies using the difference of squares: \[ \sin^2 \beta = 1 - \cos^2 \beta \] Since \(1 - \cos^2 \beta = \sin^2 \beta\), we have: \[ \sin^2 \beta = \sin^2 \beta \] This is true for all \(\beta\) where \(\sin \beta\) and \(\cos \beta\) are defined, indicating that the original equation holds for all angles \(\beta\) where \(\sin \beta \neq 0\) and \(\cos \beta \neq 1\). Thus, the solution is all values of \(\beta\) excluding those for which \(\sin \beta = 0\) (i.e., \( \beta = n\pi \) for any integer \( n \)) and \(\cos \beta = 1\) (i.e., \( \beta = 2n\pi \)).