Prove? #cotx/(cscx-1)=(cscx+1)/cotx# Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer MathFact-orials.blogspot.com Jan 17, 2017 Remember that #1+cot^2x=csc^2x#. Explanation: #cotx/(cscx-1)=(cscx+1)/cotx# Remember that #1+cot^2x=csc^2x#. This becomes useful if we multiply the terms with #cscx# to get them squared: #cotx/(cscx-1)((cscx+1)/(cscx+1))=(cscx+1)/cotx# #(cotxcscx+cotx)/(csc^2x-1)=(cscx+1)/cotx# We can now use #csc^2x-1=cot^2x# #(cotxcscx+cotx)/(cot^2x)=(cscx+1)/cotx# #(cscx+1)/(cotx)=(cscx+1)/cotx# 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 14530 views around the world You can reuse this answer Creative Commons License