Results 1 to 5 of 5

- November 12th 2008, 07:13 AM #1

- Joined
- Jul 2008
- From
- NYC
- Posts
- 1,489

- November 12th 2008, 07:18 AM #2

- November 12th 2008, 07:44 AM #3

- Joined
- Apr 2005
- Posts
- 17,994
- Thanks
- 2355

I prefer the way Chris did this but since he got in first (and I just

**have**to put in my oar!) here's another way:

Use trig identities. [tex]tan^2(\theta)+1= sec^2(\theta)= 1/cos^2(\theta)[/itex] so

Replacing by , that gives [tex]cos(tan^{-1}(v))= \frac{1}{\sqrt{1+ tan^2(tan^{1}(v))}= \frac{1}{\sqrt{1+ v^2}}.

- November 12th 2008, 04:43 PM #4

- Joined
- Jul 2008
- From
- NYC
- Posts
- 1,489

- November 12th 2008, 06:34 PM #5