How do you evaluate #sec(-pi/18) * cos(37pi/18)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Nghi N Mar 6, 2018 = 1 Explanation: #sec (- pi/18) = 1/(cos (-pi/18)) = 1/cos (pi/18)# #cos ((37pi)/18) = cos (pi/18 + (36pi)/18) = cos (pi/18)# The product transform itself to: #P = cos (pi/18)/(cos (pi/18)) = 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 4476 views around the world You can reuse this answer Creative Commons License