How do you evaluate #sec(25pi/6)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Alan P. Apr 19, 2016 #sec(25pi/6)=2/sqrt(3)# Explanation: #(25pi)/6 = 2xx(2pi)+pi/6# Each #2pi# represents one complete rotation about the origin. So #color(white)("XXX")(25pi)/6=pi/6# By definition #sec = "hypotenuse"/"adjacent"# So #sec((25pi)/6)=sec(pi/6)=2/sqrt(3)# 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 6339 views around the world You can reuse this answer Creative Commons License