How do you prove: #cotx-cosx=cotx(1-sinx)#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Don't Memorise Apr 16, 2015 Right Hand Side : #cotx(1-sinx)# # = cot x - cotx*sinx# (Distributive Property of Multiplication) We know that #color(blue)(cot x = cos x / sin x# # = cot x - (cos x/cancel sin x)*cancel sinx# # = cot x - cos x# is the same as the Left hand Side Hence proved. 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 6512 views around the world You can reuse this answer Creative Commons License