How do you prove #(sin x - sin y) / (cos x + cos y) + ( cos x - cos y) / ( sin x + sin y) = 0#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Bdub Apr 17, 2016 See below Explanation: Left Side=#((sinx-siny)(sinx +siny)+(cosx-cosy)(cosx+cosy))/((cosx+cosy)(sinx+siny))# #=(sin^2x-sin^2y+cos^2x-cos^2y)/((cosx+cosy)(sinx+siny))# #=((sin^2x+cos^2x)-(sin^2y+cos^2y))/((cosx+cosy)(sinx+siny))# #=(1-1)/((cosx+cosy)(sinx+siny))# #=0/((cosx+cosy)(sinx+siny))# #=0# #=#Right 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 3274 views around the world You can reuse this answer Creative Commons License