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).
 A: \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 $\sin^2 A + \cos^2 A = 1.$
A: $$
\frac{1+\sin A +\cos A}{1-\sin A + \cos A} = \frac{1+\sin A}{\cos A}\\
\cos A + \sin A \cos A + \cos^2 A = (1- \sin A + \cos A) (1+\sin A)\\
\cos A + \sin A \cos A + \cos^2 A = 1- \sin A + \cos A + \sin A - \sin^2 A + \sin A \cos A\\
\cos A + \sin A \cos A + \cos^2 A + \sin^2 A = 1- \sin A + \cos A + \sin A + \sin A \cos A\\
\cos A + \sin A \cos A + 1 = 1 + \cos A + \sin A \cos A\\
$$
Read from bottom to top.
A: Asserting that$$\frac{1+\sin A+\cos A}{1-\sin A+\cos A}=\frac{1+\sin A}{\cos A}$$is equivalent to asserting that $(1+\sin A+\cos A)\cos A=(1-\sin A+\cos A)(1+\sin A)$. But$$(1+\sin A+\cos A)\cos A=\cos A+\sin(A)\cos(A)+\cos^2A$$and\begin{align}(1-\sin A+\cos A)(1+\sin A)&=(1-\sin A)(1+\sin A)+\cos A+\cos(A)\sin(A)\\&=1-\sin^2A+\cos A+\cos(A)\sin(A)\\&=\cos^2A+\cos A+\cos(A)\sin(A).\end{align}
