(c) \( \cos ^{2} A\left[\frac{1}{\sin A-1}+\frac{1}{\sin A+1}\right]=-2 \sin A \)

To delve into the equation \( \cos^{2} A\left[\frac{1}{\sin A-1}+\frac{1}{\sin A+1}\right]=-2 \sin A \), we can first simplify the expression on the left side. This part reduces to \( \cos^{2} A \cdot \frac{(\sin A+1) + (\sin A-1)}{(\sin A-1)(\sin A+1)} = \cos^{2} A \cdot \frac{2\sin A}{1 - \sin^2 A} = \cos^{2} A \cdot \frac{2\sin A}{\cos^{2} A} = 2 \sin A \). Now we have \( 2 \sin A = -2 \sin A \), which leads to \( 4 \sin A = 0 \). This means \( \sin A = 0 \). Consequently, the solutions to this equation are \( A = n\pi \) for \( n \in \mathbb{Z} \). To visualize this, consider plotting the sine function; it crosses the x-axis (where \( \sin A = 0 \)) at integers multiples of \( \pi \), radiating outwards like ripples in a pond!