4) \( \frac{\csc ^{2} A-1}{\cot A \cdot \cos A}=\frac{1}{\operatorname{sen} A} \)

To solve the equation \[ \frac{\csc^2 A - 1}{\cot A \cdot \cos A} = \frac{1}{\sin A}, \] we start by recalling some trigonometric identities: 1. \(\csc A = \frac{1}{\sin A}\), so \(\csc^2 A = \frac{1}{\sin^2 A}\). 2. \(\cot A = \frac{\cos A}{\sin A}\). Using these identities, we can rewrite \(\csc^2 A - 1\): \[ \csc^2 A - 1 = \frac{1}{\sin^2 A} - 1 = \frac{1 - \sin^2 A}{\sin^2 A} = \frac{\cos^2 A}{\sin^2 A}. \] Now substituting this into the left-hand side of the equation: \[ \frac{\csc^2 A - 1}{\cot A \cdot \cos A} = \frac{\frac{\cos^2 A}{\sin^2 A}}{\cot A \cdot \cos A}. \] Next, we substitute \(\cot A\): \[ \cot A = \frac{\cos A}{\sin A}, \] thus, \[ \cot A \cdot \cos A = \frac{\cos A}{\sin A} \cdot \cos A = \frac{\cos^2 A}{\sin A}. \] Now we can rewrite the left-hand side: \[ \frac{\frac{\cos^2 A}{\sin^2 A}}{\frac{\cos^2 A}{\sin A}} = \frac{\cos^2 A}{\sin^2 A} \cdot \frac{\sin A}{\cos^2 A} = \frac{\sin A}{\sin^2 A} = \frac{1}{\sin A}. \] Now we have: \[ \frac{1}{\sin A} = \frac{1}{\sin A}. \] This shows that both sides of the equation are equal, confirming that the original equation is true for all values of \(A\) where the expressions are defined (i.e., \(A \neq n\pi\) for \(n \in \mathbb{Z}\), where \(\sin A = 0\)). Thus, the solution is verified, and the equation holds true.