# Thread: Help evaluating this trig expression!

1. ## Help evaluating this trig expression!

$\displaystyle sec(tan^{-1} 2x)$

I think the 2x has something to do with why I can't solve this problem, but even then, I still don't remember what to do about it...

2. Originally Posted by orkdork
$\displaystyle sec(tan^{-1} 2x)$

I think the 2x has something to do with why I can't solve this problem, but even then, I still don't remember what to do about it...
Create a right triangle that satisfies $\displaystyle \tan\theta=2x$. Then find $\displaystyle \sec\theta$ using the triangle constructed.

Can you proceed?

3. I think I get it, i'll try it

4. Ok, yeah, I don't know how to find the right triangle... ._.

5. Originally Posted by orkdork
Ok, yeah, I don't know how to find the right triangle... ._.
Recall $\displaystyle \tan\theta=\dfrac{\text{opp}}{\text{adj}}$. So in the case of $\displaystyle \tan\theta=2x$, 2x is the side opposite of $\displaystyle \theta$, and 1 is the side adjacent to $\displaystyle \theta$.

As a result, the hypotenuse of the triangle is $\displaystyle \sqrt{1+4x^2}$ (verify).

Now what is $\displaystyle \sec\theta$?

6. ohh...ok that makes sense now thanks