How do you prove #tan² x(1 + cot² x) = sec² x#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Gió May 10, 2015 Consider that: #tanx=sinx/cosx# #cotx=cosx/sinx# #secx=1/(cosx)# So you get: #sin^2x/cos^2x(1+cos^2x/(sin^2x))=1/cos^2x# #cancel(sin^2x)/cancel(cos^2x)((sin^2x+cos^2x)/cancel(sin^2x))=1/cancel(cos^2x)# And: #sin^2x+cos^2x=1# 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 2821 views around the world You can reuse this answer Creative Commons License