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$\cos(\tan^{-1}(x))$

I set $\tan^{-1}(x)=\theta$, then I found the inverse, which is $\tan(\theta) = x$. That would mean that $\tan(\theta) = \frac{opp}{adj} = \frac{x}{1}$. What I need to do is find the hypotenuse. I know I have to use the Pythagorean theorem, but I don't have the hypotenuse, so it's not possible.

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If you have a right triangle, then the hypotenuse is $\sqrt{x^2+1}$. –  Sanath Mar 13 at 1:07
    
How did you get that? –  igknighton Mar 13 at 1:09
    
You have $opp.=x,adj.=1$. What do you get by Pythagoras' theorem? Also, the tag "calculus" is not appropriate - "trigonometry" would be better. –  Sanath Mar 13 at 1:10
    
Oh ok, I understand now –  igknighton Mar 13 at 1:14

1 Answer 1

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Since you know you have to use Pythagoras' theorem, by the converse of Pythagoras' theorem, the triangle must be a right triangle. Thus, since $opp.=x,adj.=1$, drawing out the triangle (if you need to), and using Pythagoras' theorem gives $$\boxed{Hypotenuse=\sqrt{x^2+1}}$$

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I understand now. Thank you –  igknighton Mar 13 at 1:14

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