How do you prove #[(sec^6)x] (secxtanx) - [(sec^4)x] (secxtanx) = [(sec^5)x] [(tan^3)x] #? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer P dilip_k Apr 1, 2016 #=Secxtanx(sec^6x-sec^4x)# #=Secxtanxsec^4x(sec^2x-1)# #=Secxtanxsec^4xtan^2x# #=Sec^5xtan^3x# 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 2213 views around the world You can reuse this answer Creative Commons License