# Trignometry.. Even Odd Properties

So we have this even/odd properties in trigonometry and we had this problem:

$\sec (-(\frac{\pi}6))$

And ok i was like first off i can change the - because of even odd properties so it becomes:

$-(\sec(\frac{\pi}6))$

So next I substitute $\sec$ to $\frac1{\cos}$ ....

$-(\frac1{\cos (\frac{\pi}6)}))$

There I get $-2$ ... but the answer is $\frac{2\sqrt{3}}{3}$
I really don't understand what I did wrong here... And I figured out that she was right.. I searched up the question online and it did give the answer? What am I doing wrong here?

You have the even/odd property of $\sec$ wrong. $$\sec(-x)=\sec(x)$$
• @Dsafds np, just a minor issue. You also evaluated $\cos\left(\frac{\pi}6\right)$ incorrectly. Mar 7, 2016 at 22:44
• @Dsafds No, it should be $\frac{\sqrt{3}}2$, as Kf-Sansoo noted. Mar 7, 2016 at 22:49
• Sorry made a mistake.... I was thinking about pie/3 .. $\frac{\pi}3$ Mar 7, 2016 at 22:50
$\sec\left(-\dfrac{\pi}{6}\right)= \dfrac{1}{\cos\left(-\dfrac{\pi}{6}\right)}= \dfrac{1}{\cos\left(\dfrac{\pi}{6}\right)}= \dfrac{1}{\dfrac{\sqrt{3}}{2}}= \dfrac{2}{\sqrt{3}}= \dfrac{2\sqrt{3}}{3}$