How do you prove #tan^-1(1/4)+tan^-1(2/9)=tan^-1(1/2)#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer P dilip_k Nov 6, 2016 #LHS=tan^-1(1/4)+tan^-1(2/9)# #=tan^-1((1/4+2/9)/(1-1/4xx2/9))# #=tan^-1(((9*1+4*2)/36)/((36-2)/36))# #=tan^-1((17/36)/((34)/36))# #=tan^-1(cancel17/cancel36xxcancel36/cancel34^2)# #=tan^-1(1/2)=RHS# Proved 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 4214 views around the world You can reuse this answer Creative Commons License