How do you find the value of #sec^2(225^circ)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Nityananda Apr 28, 2017 #1/4# Explanation: #sec^2(225 ^circ) = sec^2(2xx90 + 45)^circ# #rArr [-sec45^circ]^2# Note:- [#225^circ# is in 3rd Quadrant where only tan & cot positive.] #rArr [-1/(sqrt 2)]^2# #rArr 1/4# 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 1747 views around the world You can reuse this answer Creative Commons License