# Math Help - first expression of the second

1. ## first expression of the second

Hello,

Here's the instructions:
Write the first expression in terms of the second if the terminal point determined by t is in the given quadrant.

Problem:

sec^2(t)sin^2(t), cos (t) ; any quadrant

Thanks!
-Mark

2. Use the fact that sec(x) = 1/cos(x) and that sin^2(x) + cos^2(x) = 1.

3. if that's the case, would the answer be

sec^2 = 1/sqrt(1-sin^2)

4. Originally Posted by l flipboi l
Hello,

Here's the instructions:
Write the first expression in terms of the second if the terminal point determined by t is in the given quadrant.

Problem:

sec^2(t)sin^2(t), cos (t) ; any quadrant

Thanks!
-Mark
Originally Posted by l flipboi l
if that's the case, would the answer be

sec^2 = 1/sqrt(1-sin^2)
Hi I flipboi l,

Using what Stapel told you....

$\sec^2(t) \sin^2(t)=\frac{1}{\cos^2(t)}\cdot 1-\cos^2(t)=\frac{1-\cos^2(t)}{\cos^2(t)}=\frac{1}{\cos^2(t)}-1$

5. Great!

I was wondering though, for the sec^2 (t) sin^2 (t)...why is there a sin^2 (t) there? what's it's purpose?

6. Originally Posted by l flipboi l
Great!

I was wondering though, for the sec^2 (t) sin^2 (t)...why is there a sin^2 (t) there? what's it's purpose?
I have no idea. Where did the expression originate? Is it an exercise in a textbook? Looks like this is an exercise in manipulating trigonometric expressions. Good practice, though.

7. Originally Posted by masters
I have no idea. Where did the expression originate? Is it an exercise in a textbook? Looks like this is an exercise in manipulating trigonometric expressions. Good practice, though.
yeah it came from the book. I'm trying to understand what sin^2(t) has to do with sec^2 (t) and why they're together.