How do you verify #(1-sin)/(1+sin) = (sec-tan)^2#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Bdub Oct 12, 2016 see below Explanation: #(1-sinx)/(1+sinx) = (secx-tanx)^2# Left Side : #=(1-sinx)/(1+sinx) # #=(1-sinx)/(1+sinx) * (1-sinx)/(1-sinx) # #=(1-2sinx+sin^2x)/(1-sin^2x)# #=(1-2sinx+sin^2x)/cos^2x# #=1/cos^2x-(2sinx)/cos^2x+sin^2x/cos^2x# #=1/cos^2x-2 * 1/cosx sinx/cosx+sin^2x/cos^2x# #=sec^2x-2secxtanx+tan^2x# #=(secx-tanx)(secx-tanx)# #=(secx-tanx)^2# #:.=# 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 43423 views around the world You can reuse this answer Creative Commons License