1a) \( \tan ^{2} A \cdot \cos ^{2} A+\frac{\sin ^{2} A}{\tan ^{2} A}=1 \)

To solve the equation \( \tan^2 A \cdot \cos^2 A + \frac{\sin^2 A}{\tan^2 A} = 1 \), we can rewrite \( \tan A \) as \( \frac{\sin A}{\cos A} \). Thus, \( \tan^2 A = \frac{\sin^2 A}{\cos^2 A} \). Substituting this into the equation gives: \[ \frac{\sin^2 A}{\cos^2 A} \cdot \cos^2 A + \frac{\sin^2 A}{\frac{\sin^2 A}{\cos^2 A}} = 1 \] The first term simplifies to \( \sin^2 A \), and the second term simplifies to \( \cos^2 A \): \[ \sin^2 A + \cos^2 A = 1 \] This is the fundamental Pythagorean identity, confirming that the original equation holds true for all values of \( A \) where \( \sin A \) and \( \cos A \) are defined. Thus, the equation is valid for any angle \( A \) in its domain.