How do you verify #sec^4x-sec^2x=tan^4x+tan^2x#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Nghi N. · Noah G Nov 7, 2015 Verify #sec^4 x - sec^2 x = tan ^4 x + tan^2 x# Explanation: Left side #->sec^4 x - sec^2 x # #= 1/(cos^4 x) - 1/(cos^2 x)# #= ( 1 - cos^2 x)/(cos^4 x) # #= sin^2 x/(cos^4 x)# #= tan^2 x(1/(cos^2 x))# Apply the trig identity: #1/(cos^2 x) = (1 + tan^2 x)#, we get: Left side #-> tan^2 x(1 + tan^2 x)= tan^4 x + tan^2 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 74705 views around the world You can reuse this answer Creative Commons License