How do you simplify #secxsin(pi/2-x)#? Trigonometry Right Triangles Relating Trigonometric Functions 1 Answer Nghi N. Oct 17, 2015 Simplify: #sec x.sin (pi/2 - x)# Ans: 1 Explanation: Trig unit circle --># sin (pi/2 - x) = cos x# #sec x.sin (pi/2 - x) = (1/cos x)(cos x) = 1# 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 3557 views around the world You can reuse this answer Creative Commons License