Question #7efd9 Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Bdub Feb 7, 2017 see below Explanation: Left Hand Side: #(tanx-cotx)/(tan^2x-cot^2x) = (tanx-cot x)/((tanx-cotx)(tanx+cotx))# #=(cancel(tanx-cot x) )/(cancel((tanx-cotx))(tanx+cotx))# #=1/(tanx+cotx)# #=1/(sinx/cos x +cosx/sinx)# #=1/((sin^2x+cos^2x)/(sinx cosx)# #=1/(1/(sinx cosx))# #=1* (sinxcosx)/1# #=sinxcosx# #:.=# Right Hand Side 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 1287 views around the world You can reuse this answer Creative Commons License