1. Simplify $\frac{\tan A (\sec A - \cos A)}{\sin A}$. Recall the identities: $\tan A = \frac{\sin A}{\cos A}$ and $\sec A = \frac{1}{\cos A}$. Rewrite the expression: $$\frac{\frac{\sin A}{\cos A} \left( \frac{1}{\cos A} - \cos A \right)}{\sin A} = \frac{\sin A}{\cos A} \cdot \frac{\frac{1}{\cos A} - \cos A}{\sin A}$$ Simplify inside the parentheses: $$\frac{1}{\cos A} - \cos A = \frac{1 - \cos^2 A}{\cos A}$$ Use the Pythagorean identity $1 - \cos^2 A = \sin^2 A$: $$\frac{\sin^2 A}{\cos A}$$ Substitute back: $$\frac{\sin A}{\cos A} \cdot \frac{\sin^2 A}{\cos A \sin A}$$ Cancel $\sin A$ in numerator and denominator: $$\frac{\sin A}{\cos A} \cdot \frac{\sin^2 A}{\cos A \cancel{\sin A}} = \frac{\sin A}{\cos A} \cdot \frac{\sin A}{\cos A} = \frac{\sin^2 A}{\cos^2 A}$$ Recognize this as $\tan^2 A$. **Final answer:** $$\boxed{\tan^2 A}$$