c. \( \frac{\sin 2 A}{1-\cos 2 A}=\cot A \)

To solve the equation \( \frac{\sin 2A}{1 - \cos 2A} = \cot A \), we can start by using known trigonometric identities. Recall that \( \sin 2A = 2 \sin A \cos A \) and \( \cos 2A = 1 - 2\sin^2 A \). This transforms our equation into: \[ \frac{2 \sin A \cos A}{1 - (1 - 2 \sin^2 A)} = \cot A \] This simplifies to: \[ \frac{2 \sin A \cos A}{2 \sin^2 A} = \cot A \] Cancelling out \(2\) results in: \[ \frac{\cos A}{\sin A} = \cot A \] Since \( \frac{\cos A}{\sin A} = \cot A \) is indeed an identity, the original equation holds true for all values of \(A\) where \( \sin A \neq 0 \). Thus, any angle \(A\) that does not lead to division by zero in sine is a solution! All angles \( A \) where \( \sin A \neq 0 \) satisfy this equation, leading us to conclude it's valid throughout these ranges.