Given #sinx=sqrt3/2, cosx=-1/2#, how do you find the remaining trigonometric functions? Trigonometry Right Triangles Relating Trigonometric Functions 1 Answer Gerardina C. Jan 26, 2017 #tanx=-sqrt3# #secx=-2# #cscx=2/3sqrt3# #cotx=-sqrt3/3# Explanation: Since #tantheta=sintheta/costheta#, you get #tanx=(sqrt3/cancel2)/(-1/cancel2)=-sqrt3# Since #sectheta=1/costheta#, you get #secx=1/(-1/2)=-2# Since #csctheta=1/sintheta#, you get #cscx=1/(sqrt3/2)=2/sqrt3=2/3sqrt3# Since #cottheta=1/tantheta#, you get #cotx=1/-sqrt3=-sqrt3/3# 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 7628 views around the world You can reuse this answer Creative Commons License