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 $$\frac{\cos^2 A}{1 - \sin A} = 1 + \sin A?$$ 

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
$$\dfrac {\cos^2 A}{1-\sin A}$$
$$\dfrac {1-\sin^2 A}{1-\sin A}$$
$$\dfrac {(1-\sin A)(1+\sin A)}{1-\sin A}$$
$${1+\sin A}$$

