# Math Help - Reference Triangle for Inverse Trig Functions

1. ## Reference Triangle for Inverse Trig Functions

I am not understanding how to do these problems

I have to do this for

a) sec^-1 (- sqroot 2)

b) sec^-1 (2/sq root3)

c) sec^-1 (-2)

Can someone please explain to me what to do and how to figure these problems out. Thanks!!

2. Originally Posted by KarlosK
I am not understanding how to do these problems

I have to do this for

a) sec^-1 (- sqroot 2)

b) sec^-1 (2/sq root3)

c) sec^-1 (-2)

Can someone please explain to me what to do and how to figure these problems out. Thanks!!
a) is just $\cos -\sqrt{2}$

b and c are just the cos of the term since $\sec =\frac{1}{\cos }$

Therefore, the inverse is just cosine