How do you prove #sinx+cosx=(2sin^2x-1)/(sinx-cosx)#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer GiĆ³ Apr 30, 2015 Try this: #sinx+cosx=[2(1-cos^2x)-sin^2x-cos^2x]/(sinx-cosx)# #(sinx+cosx)(sinx-cosx)=2-2cos^2x-sin^2x-cos^2x# #sin^2x-cancel(cos^2x)=2-2cos^2x-sin^2x-cancel(cos^2x)# #2sin^2x=2(1-cos^2x)# #2sin^2x=2sin^2x# Where I used the fact that: #sin^2x+cos^2x=1# 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 7034 views around the world You can reuse this answer Creative Commons License