How do you prove #(cosx/(1+sinx))+((1+sinx)/cosx)=2secx#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Alan P. · John D. Jul 23, 2015 Convert the left side into terms with common denominator and add (converting #cos^2+sin^2# to #1# along the way); simplify and refer to definition of #sec = 1/cos# Explanation: #(cos(x)/(1+sin(x)))+((1+sin(x))/cos(x))# #= (cos^2(x) + 1+2sin(x) + sin^2(x))/(cos(x)( 1+sin(x)# #= (2 +2sin(x))/(cos(x)(1+sin(x))# #= 2/cos(x)# #= 2* 1/cos(x)# #= 2sec(x)# 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 81641 views around the world You can reuse this answer Creative Commons License