How do you express #sec(-1290^circ)# as a trig function of an angle in Quadrant I? Trigonometry Right Triangles Relating Trigonometric Functions 1 Answer A. S. Adikesavan Aug 9, 2018 #- sec 30^o = - 2/ sqrt3 = -1.1547#, nearly. Explanation: #sec ( - 1290^o ) = sec ( - 7 ( 180^o ) - 30^o ) # #rArr #- 1290^o in Q_2#, and so, #sec ( - 1290^o ) = - sec ( - 30^o ) = - sec 30^o = - 2/ sqrt3# #= -1.1547#, nearly. Answer link Related questions What does it mean to find the sign of a trigonometric function and how do you find it? What are the reciprocal identities of trigonometric functions? What are the quotient identities for a trigonometric functions? What are the cofunction identities and reflection properties for trigonometric functions? What is the pythagorean identity? If #sec theta = 4#, how do you use the reciprocal identity to find #cos theta#? How do you find the domain and range of sine, cosine, and tangent? What quadrant does #cot 325^@# lie in and what is the sign? How do you use use quotient identities to explain why the tangent and cotangent function have... How do you show that #1+tan^2 theta = sec ^2 theta#? See all questions in Relating Trigonometric Functions Impact of this question 2529 views around the world You can reuse this answer Creative Commons License