How do you prove #(Sec x/Sin x)-(Sin x/Cos x)=Cot x#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Narad T. Jul 27, 2018 See the proof below Explanation: We need #sin^2x+cos^2x=1# #secx=1/cosx# #sin^2x+cos^2x=1# Therefore, #LHS=(secx/sinx)-(sinx/cosx)# #=(1/(cosxsinx))-(sinx/cosx)# #=(1-sin^2x)/(cosxsinx)# #=cos^2x/(cosxsinx)# #=cosx/sinx# #=cotx# #=RHS# #QED# 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 8865 views around the world You can reuse this answer Creative Commons License