Simplify #(1+cos2A+sin2A)/(1-cos2A+sin2A)#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Shwetank Mauria Sep 18, 2016 Please see below. Explanation: #(1+cos2A+sin2A)/(1-cos2A+sin2A)# Now #sin2A=2sinAcosA# and #cos2A=2cos^2A-1=1-2sin^2A# Hence #(1+cos2A+sin2A)/(1-cos2A+sin2A)# = #(1+2cos^2A-1+2sinAcosA)/(1-(1-2sin^2A)+2sinAcosA)# = #(1+2cos^2A-1+2sinAcosA)/(1-1+2sin^2A+2sinAcosA)# = #(2cos^2A+2sinAcosA)/(2sin^2A+2sinAcosA)# = #(2cosA(cosA+sinA))/(2sinA(sinA+cosA))# = #(2cosA)/(2sinA)# = #cotA# 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 8859 views around the world You can reuse this answer Creative Commons License