How do you verify #(sinA-cosA)^2=1-2sin^2AcotA#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Bdub Nov 12, 2016 see below Explanation: #(sinA-cosA)^2=1-2sin^2AcotA# #(sinA-cosA)(sinA-cosA)=1-2sin^2A*cosA/sinA# #sin^2A-2sinAcosA+cos^2A=1-2sin^cancel2AcosA/cancelsinA# #sin^2A+cos^2A-2sinAcosA=1- 2 sinA cos A# #1-2 sin A cos A=1- 2 sinA cos A# #:.#Left Hand Side = Right Hand Side 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 5865 views around the world You can reuse this answer Creative Commons License