How do I find the value of sec 11pi / 6? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Nghi N. Oct 18, 2015 Find #sec ((11pi)/6)# Ans: #(2sqrt3)/3# Explanation: #sec ((11pi)/6) = 1/cos ((11pi)/6)#. Find #cos ((11pi)/6)# On the trig unit circle, #cos ((11pi)/6) = cos (-pi/6 + 2pi) = cos (-pi/6) = cos (pi/6) = sqrt3/2# Finally, #sec ((11pi)/6) = 2/sqrt3 = (2sqrt3)/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 39943 views around the world You can reuse this answer Creative Commons License