$\frac{\tan A+\sin A}{\tan A-\sin A}=\frac{\sec A+1}{\sec A–1}$

$$\begin{gathered} \frac{\tan A+\sin A}{\tan A-\sin A} \\ =\frac{{\displaystyle \frac{\sin A}{\cos A}}+\sin A}{{\displaystyle \frac{\sin A}{\cos A}}-\sin A} \\ =\frac{\sin A\left({\displaystyle \frac{1}{\cos A}}+1\right)}{\sin A\left({\displaystyle \frac{1}{\cos A}}-1\right)} \\ =\frac{\sec A+1}{\sec A-1}\left(\sec A=\frac{1}{\cos A}\right) \end{gathered}$$