How do you prove #(csc x - cot x)^2 = (1 - cos x)/(1+cosx)#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Nghi N. Dec 8, 2015 Prove trig expression Explanation: #[1/(sin x) - cos x/(sin x)]^2 = (1 - cos x)^2/sin^2 x =# #= (1 - cos x)^2/(1 - cos^2 x) = (1 - cos x)^2/((1 - cos x)(1 + cos x))# = #= (1 - cos x)/(1 + cos x)# 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 2249 views around the world You can reuse this answer Creative Commons License