How do you evaluate #sec(-pi)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer georgef Aug 21, 2016 #sec(-pi)=1/cos(-pi)# Explanation: The #cos# function is an even function, and that means that #cos(-x)=cos x#, for all values of #x#. In particular #cos(-pi)=cos pi = -1#. Then, since #sec (-pi) = 1/cos(-pi)=1/(-1)=-1# Answer link Related questions How do you find the trigonometric functions of any angle? What is the reference angle? How do you use the ordered pairs on a unit circle to evaluate a trigonometric function of any angle? What is the reference angle for #140^\circ#? How do you find the value of #cot 300^@#? What is the value of #sin -45^@#? How do you find the trigonometric functions of values that are greater than #360^@#? How do you use the reference angles to find #sin210cos330-tan 135#? How do you know if #sin 30 = sin 150#? How do you show that #(costheta)(sectheta) = 1# if #theta=pi/4#? See all questions in Trigonometric Functions of Any Angle Impact of this question 2370 views around the world You can reuse this answer Creative Commons License