How do you prove #tan(2theta)=2/(cottheta-tantheta)#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Bdub Oct 7, 2016 see below Explanation: #tan 2 theta = 2/(cot theta-tan theta)# Right Side :#=2/(cot theta-tan theta)# #=2/(costheta/sin theta - sin theta/cos theta)# #=2/((cos^2theta-sin^2theta)/(sin theta cos theta)# #=2 * (sin theta cos theta)/(cos^2theta-sin^2theta)# #=(2 sin theta cos theta)/(cos^2theta-sin^2theta)# #=sin(2theta)/cos(2theta)# #=tan ( 2 theta)# #:. = # Left 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 4463 views around the world You can reuse this answer Creative Commons License