How do you verify #(secx - cscx)/(sin^2 x - cos^2 x) = (secx*cscx)/(cos(-x)-sin( -x))#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Sonnhard May 26, 2018 Use #sin^2(x)-cos^2(x)=(sin(x)-cos(x))(sin(x)+cos(x))# Explanation: we have #((sin(x)-cos(x))/(cos(x)sin(x)))/((sin(x)-cos(x))(sin(x)+cos(x))# That is #1/((sin(x)cos(x))*(sin(x)+cos(x)))=(sec(x)*csc(x))/(sin(x)+cos(x))# for #sin^2(x)ne cos^2(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 1686 views around the world You can reuse this answer Creative Commons License