$\dfrac{\tan A}{\sec A + 1} - \dfrac{\sec A - 1}{\tan A} = ?$

Using $\sec^2 A - 1 = \tan^2 A$, multiply the first fraction top and bottom appropriately: $\dfrac{\tan A}{\sec A + 1} = \dfrac{\tan A(\sec A - 1)}{\sec^2 A - 1} = \dfrac{\sec A - 1}{\tan A}$. The two terms are identical, so their difference is $0$.