How do you prove #tan2x + sec2x = (cosx + sinx) / (cosx – sinx)#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer P dilip_k Jul 1, 2016 As below Explanation: #RHS = (cosx + sinx) / (cosx – sinx)# # =( (cosx + sinx) (cosx +sinx) )/( (cosx - sinx) (cosx +sinx))# #= (cosx + sinx) ^2/ (cos^2x – sin^2x)# # = (cos^2x + sin^2x+2sinxcosx) / (cos2x)# #=(1+sin2x)/(cos2x)=1/(cos2x)+(sin2x)/(cos2x)# #=tan2x+sec2x=LHS# proved 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 14806 views around the world You can reuse this answer Creative Commons License