We need to prove
[tex]\frac{\tan A+\sin A}{\tan A-\sin A}=\frac{\sec A+1}{\sec A-1}[/tex]
[tex]\text{LHS}=\frac{\tan A+\sin A}{\tan A-\sin A}\\\\=\frac{\frac{\sin A}{\cos A}+\sin A}{\frac{\sin A}{\cos A}-\sin A}\\\\=\frac{\sin A+\sin A \cos A}{\sin A-\sin A \cos A}\\\\=\frac{\sin A(1+\cos A)}{\sin A(1-\cos A)}\\\\=\frac{1+\cos A}{1-\cos A}\\\\=\frac{1+\frac{1}{\sec A}}{1-\frac{1}{\sec A}}\\\\=\frac{\sec A+1}{\sec A-1}\\\\ =\text{RHS}[/tex]
