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

Note
\begin{eqnarray}
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}\\
&=&\frac{\sin A+1-\cos A}{\sin A-1+\cos A}\cdot\frac{\sin A+1+\cos A}{\sin A+1+\cos A}\\
&=&\frac{(\sin A+1)^2-\cos^2 A}{(\sin A+\cos A)^2-1}\\
&=&\frac{\sin^2 A-\cos^2 A+2\sin A+1}{2\sin A\cos A}\\
&=&\frac{2\sin^2 A+2\sin A}{2\sin A\cos A}\\
&=&\frac{1+\sin A}{\cos A}\\
&=&RHS
\end{eqnarray}
