How do you prove #cos^2theta/(1+sintheta)=1-sintheta#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Bdub Oct 7, 2016 see below Explanation: #cos^2theta/(1+sin theta) = 1-sin theta# Left Side : #=cos^2theta/(1+sin theta)# #=(1-sin^2theta)/(1+sin theta)# #=((1-sin theta)(1+sin theta))/(1+sin theta)# #=((1-sin theta) (cancel (1+sin theta)))/(cancel(1+sin theta)# #=1-sin theta# #:.=# Right 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 13702 views around the world You can reuse this answer Creative Commons License