Cot B + cos B divided by sec B - cos B = csc B + 1 divided by tan^2 B can it be verified? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Hriman Mar 13, 2018 Yes... see below Explanation: #(Cot B + cos B)/(sec B - cos B) = (csc B + 1)/ tan^2 B# #(CotB+cosB)/(1/cosB - cos^2 B/cosB)=# #(CosB/SinB+(cosBsinB)/sinB)/(sin^2B/cosB)=# #((CosB+cosBsinB)/sinB)*(cosB)/(sin^2B)=# #((Cos^2B+cos^2BsinB)/sin^3B)=# #Cos^2B/sin^3B+(cos^2BsinB)/sin^3B=# #Cos^2B/sin^2B*1/sinB+cos^2B/sin^2B=# #cot^2B*cscB+cot^2B=# #cot^2B*cscB+cot^2B=# #cscB/tan^2B+1/tan^2B=# #(cscB+1)/tan^2B# 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 2364 views around the world You can reuse this answer Creative Commons License