How do you prove #(cosx-1)(tanx-1)=0#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer maganbhai P. Apr 12, 2018 Please check the Question. Explanation: Here, #(cosx-1)(tanx-1)=0# If we take , #x=30^circ ,#then #LHS=(cos30^circ-1)(tan30^circ-1)# #=(sqrt3/2-1)(1/sqrt3-1)# #~~(0.87-1)(0.58-1)# #~~(-0.13)(-0.42)# #~~0.055!=0# #LHS!=RHS# 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 2362 views around the world You can reuse this answer Creative Commons License