Question

Establish the identity:
1/1-(cos^2 theta)/(1+sin theta) = sin theta

Answer 1

mmmm that's difficult to read.
Lemme know if this is formatted correctly,\[\Large\rm \frac{\left(\frac{1}{1-\cos^2\theta}\right)}{1+\sin \theta}=\sin \theta\]Like that..?

Answer 2

\(\bf \cfrac{1}{1}-\cfrac{cos^2(\theta)}{1+sin(\theta)}=sin(\theta)\quad ?\)

Answer 3

Sorry about that, I'm fairly new to this site.
It's

\[1- [(\cos \theta ^{2})/(1+\sin \theta)] = \sin \theta\]

Answer 4

I hope that's somewhat better.

Answer 5

So you had it comepletely correct.

Answer 6

Mmm ok here is one approach :)
\[\Large\rm 1-\frac{\cos^2\theta}{1+\sin \theta}\color{royalblue}{\left(\frac{1-\sin \theta}{1-\sin \theta}\right)}\]
We could multiply this second term, top and bottom, by the conjugate of our denominator.

Answer 7

The reason we do this is because it will allow us to make use of our Pythagorean Identity.
Leave the top alone, multiplying out the bottom gives us:\[\Large\rm 1-\frac{\cos^2\theta(1-\sin \theta)}{1-\sin^2 \theta}\]Does the multiplication make sense?

Answer 8

Okay, that's a much better approach than mine. The multiplication makes much more sense. Thank you.

Answer 9

Oh you got it from there? ^^

Answer 10

I believe so. Thank you so much!!

Answer 11

\[1-\dfrac{cos^2(x)}{1+sin(x)}=\dfrac{1+sinx}{1+sinx}-\dfrac{cos^2(x)}{1+sin(x)}\]

replace \(cos^2 = 1 - sin^2\)
\[\dfrac{1+sinx-1+sin^2x}{1+sinx}=\dfrac{sin^2x+sinx}{1+sinx}=\dfrac{sinx(1+sinx)}{1+sinx}=sinx\]

Answer 12

Thank you, Ooops!

Answer 13

:)

Answer 14

I will tell u a easier way in 2steps