How do you prove #costhetacottheta=costheta#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer sente Oct 24, 2016 The identity is not true in general Explanation: Consider #theta = pi/3#. Then #cos(theta) = 1/2#, and #cot(theta) = cos(theta)/sin(theta) = (1/2)/(sqrt(3)/2) =sqrt(3)/3# So #cos(theta)cot(theta) = 1/2(sqrt(3)/3) = sqrt(3)/6 != 1/2 = cos(theta)# 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 986 views around the world You can reuse this answer Creative Commons License