If sec(t) = 3, how do you find the exact value of #sin^2 (t)#? Trigonometry Right Triangles Relating Trigonometric Functions 1 Answer Nghi N. Nov 27, 2015 Find #sin^2 t# Ans: #sin^2 t = 8/9# Explanation: #sec t = 1/(cos t) = 3# --> #cos t = 1/3# --> #cos^2 t = 1/9# #sin^2 t = 1 - cos^2 t = 1 - 1/9 = 9/9 - 1/9 = 8/9# 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 3009 views around the world You can reuse this answer Creative Commons License