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

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

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

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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}}.

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

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