Answer:
When proving identities, the answer is in the explanation.
Step-by-step explanation:
[tex]\frac{\cos(y)}{1-\sin(y)}[/tex]
I have two terms in this denominator here.
I also know that [tex]1-\sin^2(\theta)=\cos^2(theta)[/tex] by Pythagorean Identity.
So I don't know how comfortable you are with multiplying this denominator's conjugate on top and bottom here but that is exactly what I would do here. There will be other problems will you have to do this.
[tex]\frac{\cos(y)}{1-\sin(y)} \cdot \frac{1+\sin(y)}{1+\sin(y)}[/tex]
Big note here: When multiplying conjugates all you have to do is multiply fist and last. You do not need to do the whole foil. That is when you are multiplying something like [tex](a-b)(a+b)[/tex], the result is just [tex]a^2-b^2[/tex].
Let's do that here with our problem in the denominator.
[tex]\frac{\cos(y)}{1-\sin(y)} \cdot \frac{1+\sin(y)}{1+\sin(y)}[/tex]
[tex]\frac{\cos(y)(1+\sin(y))}{(1-\sin(y))(1+\sin(y)}[/tex]
[tex]\frac{\cos(y)(1+\sin(y))}{1^2-\sin^2(y)}[/tex]
[tex]\frac{\cos(y)(1+\sin(y))}{1-\sin^2(y)}[/tex]
[tex]\frac{\cos(y)(1+\sin(y))}{cos^2(y)}[/tex]
In that last step, I apply the Pythagorean Identity I mentioned way above.
Now You have a factor of cos(y) on top and bottom, so you can cancel them out. What we are really saying is that cos(y)/cos(y)=1.
[tex]\frac{1+\sin(y)}{cos(y)}[/tex]
This is the desired result.
We are done.
