Question #a32b3 Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Narad T. Mar 13, 2017 See proof below Explanation: We use #(a+b)^2=a^2+2ab+b^2# #sin^2x+cos^2x=1# Dividing by #cos^2x# #sin^2x/cos^2x+1=1/cos^2x# #tan^2x+1=sec^2x# #tan^2x=sec^2x-1# Therefore, #LHS=(tanx+secx)^2# #=tan^2x+sec^2x+2tanxcosx# #=sec^2x-1+sec^2x+2tanxsecx# #=2sec^2x+2tanxsecx-1# #=RHS# #QED# 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 1169 views around the world You can reuse this answer Creative Commons License