How do you evaluate #csc 75#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Ratnaker Mehta Jul 22, 2016 #csc75=(sqrt6-sqrt2)~=1.0353#.. Explanation: To find #csc75#, we first find #sin75 :-# #sin75=sin(45+30)=sin45cos30+cos45sin30# #=1/sqrt2*sqrt3/2+1/sqrt2*1/2# #=(sqrt3+1)/(2sqrt2)# #=(sqrt6+sqrt2)/4#. Hence, #csc75=1/sin75# #=4/(sqrt6+sqrt2)# #{4(sqrt6-sqrt2)}/{(sqrt6+sqrt2)(sqrt6-sqrt2)}# #=(sqrt6-sqrt2)#. Taking, #sqrt6~=2.4495, and, sqrt2~=1.4142#, we get, #csc75~=2.4495-1.4142~=1.0353#. 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 8652 views around the world You can reuse this answer Creative Commons License