How do you prove (cscx+cotx)(cscx-cotx)=1? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Narad T. Mar 23, 2018 See the proof below Explanation: We need cscx=1/sinx cotx=cosx/sinx sin^2x+cos^2x=1 Therefore, LHS=(cscx+cotx)(cscx-cotx) =csc^2x-cos^2x =1/sin^2x-cos^2x/sin^2x =(1-cos^2x)/(sin^2x) =sin^2x/sin^2x =1 =RHS QED 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 9468 views around the world You can reuse this answer Creative Commons License