Question in a trigonometric equation. I tried it a lot but am not able to get this.Pls help in how should I think when solving this type of question and which side is better to try to simplify first (LHS or RHS).Please share the solution in that way.
$$
\frac{1-\sin A}{1+\sin A} = 1 + 2 \tan A \left(\tan A - \sec A \right)
$$
One of the ways I tried but am not understand that how to simplify it in such a way that you get the RHS?

Starting from the LHS we have $$\frac{1-\sin(A)}{1+\sin(A)}$$
$$=\frac{1-\sin(A)}{1+\sin(A)}\cdot\frac{1-\sin(A)}{1-\sin(A)}=\frac{1-2\sin(A)+\sin^2(A)}{1-\sin^2(A)}$$
$$=\frac{\cos^2(A)-2\sin(A)+2\sin^2(A)}{\cos^2(A)}$$
using the identity $\sin^2(A)+\cos^2(A)=1$. Can you end it now?
