Proving trigonometric identities I’ve had a bit of difficulty of this question:
(1+sinA+cosA)/(1-sinA+cosA)=(1+sinA)/cosA
I tried to do:
(SinA)^2+(CosA)^2+sinA+cosA/(SinA)^2+(CosA)^2-sinA+cosA=(1+sinA)/cosA
But then I’m kind of lost. Any help will be appreciated! Additionally, I am not allowed to move one side to another (over the equal sign).

\begin{align*} &\frac{1 + \sin A + \cos A}{1-\sin A + \cos A} = \frac{1 + \sin A}{\cos A} \\ &\iff \cos A + \sin A \cos A + \cos^2 A = 1 - \sin^2 A + \sin A \cos A + \cos A \\ &\iff \cos A + \sin A \cos A + \cos^2 A = \cos A + \sin A \cos A + \cos^2 A, \end{align*} where we used $\sin^2 A + \cos^2 A = 1.$
