How do you verify #(cscx-1)/(1-sinx) =cscx#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Shwetank Mauria Apr 13, 2016 Please see below. Explanation: #(cscx-1)/(1-sinx)# = #(1/sinx-1)/((1-sinx))# = #((1-sinx)/sinx)/((1-sinx))# = #((1-sinx))/sinx xx1/((1-sinx))# = #(cancel(1-sinx))/sinx xx1/(cancel(1-sinx))# = #1/sinx# = #cscx# 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 7067 views around the world You can reuse this answer Creative Commons License