onesun0000
Junior Member
- Joined
- Dec 18, 2018
- Messages
- 83
There are two problems with this.
1) You divided by 1 - cos(x). Thus you have the restriction that \(\displaystyle cos(x) \neq 1\).
2) I'm guessing this is along the lines of a typo: \(\displaystyle \dfrac{1}{sin(x)} = csc(x)\), not sec(x).
Notice the interesting little feature of this. Since csc(0) is undefined, and the only place that cos(x) = 1 is at x = 0, we actually don't need 1). But it's always good to check anyway.
-Dan
We rather you check. Plug some angles into both sides and see if the results are equal? Try 3-4 angles.oh my gosh. yes. sorry. so is sin (x/2) csc x correct?