How do you verify #(1+cos2x)/(sin2x) = cotx#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Massimiliano Jul 3, 2015 Since #cos2x=cos^2x-sin^2x=1-2sin^2x=2cos^2x-1# and #sin2x=2sinxcosx# then: #(1+2cos^2x-1)/(2sinxcosx)=cotxrArr# #(2cos^2x)/(2sinxcosx)=cotxrArr# #cosx/sinx=cotxrArr# #cotx=cotx#. 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 30767 views around the world You can reuse this answer Creative Commons License