How do you evaluate #sec(60)+csc^2(pi/3)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Gerardina C. Aug 3, 2016 #10/3# Explanation: Since #secalpha=1/cosalpha# and #cscalpha=1/sinalpha# the given expression becomes: #1/cos 60+1/sin^2(pi/3)# Since #cos 60=1/2# and #sin (pi/3)=sqrt(3)/2#, #2+4/3=10/3# 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 1982 views around the world You can reuse this answer Creative Commons License