Prove that #cosB/(1-sinB)=secB+tanB#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Shwetank Mauria Sep 10, 2016 Please see below. Explanation: #cosB/(1-sinB)# = #cosB/(1-sinB)xx(1+sinB)/(1+sinB)# = #(cosB(1+sinB))/(1-sin^2B)# = #(cosB(1+sinB))/cos^2B# = #(1+sinB)/cosB# = #1/cosB+sinB/cosB# = #secB+tanB# Answer link Related questions What does it mean to prove a trigonometric identity? How do you prove #\csc \theta \times \tan \theta = \sec \theta#? How do you prove #(1-\cos^2 x)(1+\cot^2 x) = 1#? How do you show that #2 \sin x \cos x = \sin 2x#? is true for #(5pi)/6#? How do you prove that #sec xcot x = csc x#? How do you prove that #cos 2x(1 + tan 2x) = 1#? How do you prove that #(2sinx)/[secx(cos4x-sin4x)]=tan2x#? How do you verify the identity: #-cotx =(sin3x+sinx)/(cos3x-cosx)#? How do you prove that #(tanx+cosx)/(1+sinx)=secx#? How do you prove the identity #(sinx - cosx)/(sinx + cosx) = (2sin^2x-1)/(1+2sinxcosx)#? See all questions in Proving Identities Impact of this question 8889 views around the world You can reuse this answer Creative Commons License