How do you simplify #cos^2B / (1-sinB)#? Trigonometry Right Triangles Relating Trigonometric Functions 1 Answer Alan P. Oct 24, 2015 #(cos^2(B))/(1-sin(B)) = 1+sin(B)# Explanation: Based on the definitions of #sin# and #cos# and on the Pythagorean Theorem #color(white)("XXX")cos^2(B)+sin^2(B) = 1# or #color(white)("XXX")cos^2(B) = 1-sin^2(B)# #color(white)("XXXXXXX")=(1-sin(B))*(1+sin(B))# Therefore #color(white)("XXX")(cos^2(B))/(1-sin(B)) = (cancel((1-sin(B)))*(1+sin(B)))/(cancel((1-sin(B)))# 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 3890 views around the world You can reuse this answer Creative Commons License