What is #(1+cosx)/(1+secx)# for #x=pi/3#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Shwetank Mauria Jan 2, 2017 #(1+cosx)/(1+secx)=cosx# and for #x=pi/3#, it is #1/2#. Explanation: #(1+cosx)/(1+secx)# = #(1+cosx)/(1+1/cosx)# = #(1+cosx)/((cosx+1)/cosx)# = #(1+cosx)xxcosx/(1+cosx)# = #cosx# Hence #(1+cosx)/(1+secx)=cosx# for all values of #x# and for #x=pi/3# #(1+cosx)/(1+secx)=(1+cos(pi/3))/(1+sec(pi/3))# = #(1+1/2)/(1+2)=(3/2)/3=3/2xx1/3=1/2# As #cosx=cos(pi/3)=1/2# for #x=pi/3#, #(1+cosx)/(1+secx)=cosx# 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 1121 views around the world You can reuse this answer Creative Commons License