How do you prove that #(2sinx)/[secx(cos4x-sin4x)]=tan2x#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Kalyanam S. May 28, 2018 As proved Explanation: I suppose the sum is #(2 sin x) / (sec x (cos^4 x - sin ^4 x)) = tan 2x# #=> (2 sin x cos x) / ((cos^2 x + sin-^2x)* (cos ^2x - sin^2x))# #color(crimson)(sin 2x = 2 sin x cos x#, identity #color(crimson)(cos^2 x + sin ^2 x -= 1#, identity #color(crimson)(cos^2x - sin^2 x = cos 2x#, identity #:. => (sin 2x) / (cos 2x) = tan 2x = R H S# 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 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)#? How do you prove the identity #tan^2x/(secx+1)= (1-cosx)/cosx#? See all questions in Proving Identities Impact of this question 19983 views around the world You can reuse this answer Creative Commons License