<s>[INST] Prove $\frac{\cos^2 A}{1 - \sin A} = 1 + \sin A$ by the Pythagorean theorem. How do I use the Pythagorean Theorem to prove  that $$\frac{\cos^2 A}{1 - \sin A} = 1 + \sin A?$$ 
 [/INST] Let's simplify matters:
In a triangle with hypotenuse equal to $1$ (think of the unit circle, an angle $A$ between the $x$ axis and the hypotenuse, we know that $$\sin A=\frac{\text{opposite}}{\text{hypotenuse}}\quad \text{and}\quad \cos A=\frac{\text{adjacent}}{\text{hypotenuse}}$$ then, since hypotenuse $=1$, we have the leg opposite the angle $A$ given by $\sin A=\text{opposite}/1$ and the leg along the x-axis of length $\cos A=\text{adjacent}/1$.  Now, by the Pythagorean Theorem, and substitution, we have that
$$\begin{align}\text{opposite}^2 + \text{adjacent}^2 &= \text{hypotenuse}^2 = 1^2 \\ \sin^2A +\cos^2 A & = 1\end{align}$$
This gives us the well-known identity: $$\sin^2A + \cos^2 A = 1\tag{1}$$
We can express this identity in terms of $\cos^2 A$ by subtracting $\sin^2 A$ from both sides of the identity to get $$\begin{align}\cos^2 A & = 1 - \sin^2 A  \\ &= (1)^2 - (\sin A)^2\tag{2}\end{align}$$
Now, we know that for any difference of squares, we can factor as follows: $$(x^2 - y^2) = (x +y)(x - y)\tag{3}$$
Since equation $(2)$ is a difference of squares, we have that $$\begin{align}\cos^2 A &= 1 - \sin^2 A \\  & = (1)^2 - (\sin A)^2 \\  &= (1 +\sin A)(1 - \sin A)\tag{4}\end{align}$$
Substituting gives us:
$$\begin{align}\frac{\cos^2 A}{1 -\sin A} & = \frac{1 - \sin^2 A}{1 -\sin A} \\ \\ &= \frac{(1 + \sin A)(\color{blue}{\bf 1 - \sin A})}{\color{blue}{\bf 1 -\sin A}}\\ \\ & = 1 + \sin A\end{align}$$
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