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

-
if i multiply it would it show that A is an acute angle? i have a given figure here and its a right triangle –  user83562 Jun 23 '13 at 14:25
Do you know the identity $\cos^2 A + \sin^2 A = 1$? –  Javier Badia Jun 23 '13 at 14:58

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

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

-
why don't you make things precise : $$1=\frac{c^2}{b^2}+\frac{a^2}{b^2}=\cos^2A+\sin^2A$$ $$\implies \cos^2A=1-\sin^2A=(1-\sin A)(1+\sin A)$$ $$\implies \frac{\cos^2A}{1-\sin A}=1+\sin A$$ –  lab bhattacharjee Jun 23 '13 at 14:32
how did it become 1-sin^2 A? –  user83562 Jun 23 '13 at 14:49
@user83562: Did you read lab's (excellent) comment? Since $$1=\cos^2A+\sin^2A,$$ then $$\cos^2A=1-\sin^2A.$$ By difference of squares formula, we have $$\cos^2A=(1-\sin A)(1+\sin A).$$ From there, it's almost immediate. –  Cameron Buie Jun 23 '13 at 15:18

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 $$\mathrm{opposite}^2 + \mathrm{adjacent}^2 = 1^2.$$

-

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}

-
You begin with "Given the well known identity,....", but the questions seems to be how to get that well known identity from the Pythagorean theorem. –  Michael Hardy Jun 23 '13 at 16:14
@amWhy: You are always welcome my friend! Time for another cheesy movie (trying to add other activities in my life). ;-) –  Amzoti Jun 24 '13 at 2:47

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)

-
not sure how much this derivation from the Left hand side to the Right is required, as this is often not the natural derivation. For example, please find my comment in the other answer. –  lab bhattacharjee Jun 23 '13 at 14:38