Prove that #(tanx+tany)/(tanx-tany)=2cscx#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Shwetank Mauria Mar 26, 2017 #(tanx+tany)/(tanx-tany)=-cos(x-y)/cos(x+y)# not #2cscx# Explanation: #(tanx+tany)/(tanx-tany)# = #(sinx/cosx+cosy/siny)/(sinx/cosx-cosy/siny)# = #((sinxsiny+cosxcosy)/(cosxsiny))/((sinxsiny-cosxcosy)/(cosxsiny))# = #(cosxcosy+sinxsiny)/-(cosxcosy-sinxsiny)# = #-cos(x-y)/cos(x+y)# Hence #(tanx+tany)/(tanx-tany)=-cos(x-y)/cos(x+y)# not #2cscx# 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 3566 views around the world You can reuse this answer Creative Commons License