How do you find the exact value of #sec((3pi)/2)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer bp Jan 25, 2017 #oo# Explanation: Since #2pi# means #360^o# means #0^o#, subtract #2pi# from #(3pi)/2#. Thus #sec ((3pi)/2)# = #sec ((3pi)/2-2pi) = sec (-pi/2)= sec (pi/2) = 1/(cos(pi/2)# # = 1/0 = oo# 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 7944 views around the world You can reuse this answer Creative Commons License