How do you prove #tan^4 x + 2tan^2 x + 1 = sec^4 x#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer sjc Jun 12, 2018 see below Explanation: we need the identity #color(red)(1+tan^2x=sec^2x)# #LHSrarrtan^4x+2tan^2x+1# #=tan^4x+tan^2x+color(red)(tan^2x+1)# #=tan^4x+tan^2x+sec^2x# #=tan^2xcolor(red)((tan^2x+1))+sec^2x# #=tan^2x(sec^2x)+sec^2x# #=sec^2xcolor(red)((tan^2x+1))# #=sec^2xsec^2x# #=sec^4x# 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 16565 views around the world You can reuse this answer Creative Commons License