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 $$\frac{\cos^2 A}{1 - \sin A} = 1 + \sin A?$$ 
 A: suppose there is a right angle tri. having b is hypo.,c is base ,a is perpendicular
than $\cos A=\frac cb$ and $\sin A=\frac ab$
using pythgo. theo. $\;b^2=a^2+c^2$
$$\dfrac {\cos^2 A}{1-\sin A}$$
$$\dfrac {\frac {c^2}{b^2}}{1-\frac {a}{b}}$$
$$\dfrac {\frac {c^2}{b^2}}{\frac {b-a}{b}}$$
$$\dfrac {c^2b}{b^2(b-a)}$$
$$\dfrac {c^2}{b(b-a)}$$
$$\dfrac {b^2-a^2}{b(b-a)}$$
$$\dfrac {b+a}{b}$$
$$1+\dfrac {a}{b}$$
$$1+\sin A$$
alternatively
$$\dfrac {\cos^2 A}{1-\sin A}$$
$$\dfrac {1-\sin^2 A}{1-\sin A}$$
$$\dfrac {(1-\sin A)(1+\sin A)}{1-\sin A}$$
$${1+\sin A}$$

A: 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:
$$\begin{align}\frac{\cos^2 A}{1 -\sin A} & = \frac{1 - \sin^2 A}{1 -\sin A} \\ \\ &= \frac{(1 + \sin A)(\color{blue}{\bf 1 - \sin A})}{\color{blue}{\bf 1 -\sin A}}\\ \\ & = 1 + \sin A\end{align}$$
A: If you allow yourself some simple algebra, then you see that
$$
\frac{\cos^2 A}{1-\sin A} = 1+\sin A\text{ is equivalent to } \cos^2 A = (1+\sin A)(1-\sin A)
$$
and that last expression is equal to $1-\sin^2 A$.  So you're asking how to prove $\cos^2 A=1-\sin^2 A$.  And that's the same as proving $\cos^2 A + \sin^2 A = 1$.
So the question is: How can you prove that $\cos^2 A + \sin^2 A = 1$ by using the Pythagorean theorem?
If you know that $\sin =\dfrac{\mathrm{opposite}}{\mathrm{hypotenuse}}$ and $\cos=\dfrac{\mathrm{adjacent}}{\mathrm{hypotenuse}}$, then this becomes easy if you consider a right triangle in which the length of the hypotenuse is $1$.  Then you have $\sin=\mathrm{opposite}$ and $\cos=\mathrm{adjacent}$.  Now the Pythagorean theorem says that
$$
\mathrm{opposite}^2 + \mathrm{adjacent}^2 = 1^2.
$$
A: Let us denote base of right angle triangle as b, perpendicular ( height ) as p, and hypotenuse as h, 
$$\cos A = \frac{b}{h} ; \qquad \sin A = \frac{p}{h}\tag{i}$$
Therefore, $$\frac{\cos^2A}{1-\sin A} = \frac{\frac{b^2}{h^2}}{1-\frac{p}{h}}$$
[By putting the values of $\cos A$ and $\sin A$ from $(i)$]
Which after simplification gives you:
$$\frac{b^2}{h(h-p)}\tag{ii}$$
Now as
$$b^2 = h^2-p^2\quad\text{[using pythagoreous theorem]}\tag{iii}$$
By putting the value of $b^2$ from (iii) in (ii) you get : 
$$\frac{h^2-p^2}{h(h-p)} = \frac{h+p}{h} = 1+ \frac{p}{h} = 1+ \sin A$$
(hence proved)
