How do you prove: # [(sinx/cosx) +(cosx/sinx)] = sinx^2+cosx^2#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer sente Mar 26, 2016 The given identity is false. Explanation: Consider #x = pi/4# #sin(pi/4)/cos(pi/4)+cos(pi/4)/sin(pi/4) = (sqrt(2)/2)/(sqrt(2)/2)+(sqrt(2)/2)/(sqrt(2)/2) = 1+1=2# but #sin((pi/4)^2)+cos((pi/4)^2) = sin(pi^2/16)+cos(pi^2/16) ~~1.394# 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 2275 views around the world You can reuse this answer Creative Commons License