I have to find the exact answer for this.

$$\sec \left( \tan^{-1} 5 \right)$$

Question

I have to find the exact answer for this.

$$\sec \left( \tan^{-1} 5 \right)$$

anonymous · Answer

|dw:1357065660623:dw|

anonymous · Answer

there is a picture of a triangle with an angle for which $\tan(\theta)=5$ 
all you are missing is the hypotenuse, which you get via pythagoras

anonymous · Answer

$$h=\sqrt{1^2+5^2}=\sqrt{26}$$ |dw:1357065774669:dw|

anonymous · Answer

since secant is hypotenuse over adjacent, you get 
$$\sec(\theta)=\frac{\sqrt{26}}{1}=\sqrt{26}$$

anonymous · Answer

if you want, redo this with 
$$\sec(\tan^{-1}(x))$$ and get a formula with an $x$ instead of the number $5$

anonymous · Answer

Wow.  So simple I couldn't see it.  I was lost because I didn't know where tan 5 was on the unit circle.  Geez.  Thanks for opening my eyes,

anonymous · Answer

yw 
easy when you know what to do right?  that is, viewed the right way.  like being asked "if the tangent is 5, what is the secant?"