How do you prove #1-cosx^2/(1+sinx)= sinx#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Anees Apr 15, 2015 Solution #1-(cos^2x)/(1+sinx)=sinx....(i)# As #cos^2x=1-sin^2x# So, L.H.S of equation no. #(i)# #=1-(1-sin^2x)/(1+sinx)# #=1-((1-sinx)(1+sinx))/((1+sinx))# #=1-((1-sinx)cancel((1+sinx)))/cancel((1+sinx))# #=1-(1-sinx)# #=1-1+sinx# #=cancel1-cancel1+sinx# #=sinx# #=R.H.S# 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 5050 views around the world You can reuse this answer Creative Commons License