Prove the following trigonometric identity: $\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{1+\sin A}{\cos A}$ [duplicate]

Prove: $$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{1+\sin A}{\cos A}$$

My attempt:

LHS= $$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}$$ $$=\frac{\frac{\sin A}{\cos A}+\frac{1}{\cos A}-1}{\frac{\sin A}{\cos A}-\frac{1}{\cos A}+1}$$ $$=\frac{\sin A+1-\cos A}{\sin A-1+\cos A}$$ $$\text{Using componendo and dividendo}$$ $$\frac{\sin A+1-\cos A+\sin A-1+\cos A}{\sin A+1-\cos A-\sin A+1-\cos A}$$ $$\frac{2\sin A}{2-2\cos A}=\frac{\sin A}{1-\cos A}=\frac{\sin A(1+\cos A)}{(1-\cos A)(1+\cos A)}$$ $$\frac{\sin A(1+\cos A)}{\sin^2 A}=\frac{1+\cos A}{\sin A}$$ $$\text{Which is not equal to right hand side!}$$

Have I done something wrong in the componendo dividendo step? I dont know how to use componendo dividendo rule . I saw it being used like this in some question and hence applied it here the same way. Maybe I am wrong in application of that rule. Please tell me the right way to use it. Thank you.