How do you prove #1/(sec D - tan D) + 1/(sec D + tan D) = 2 sec D#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Nghi N. Nov 7, 2015 Prove : #1/(sec D - tan D) + 1/(sec D + tan D) = 2sec D# Explanation: Left side --> #((sec D + tan D) + (sec D - tan D))/(sec^2D - tan^2 D) = # #= (2sec D)/1# because: #sec^2 D - tan^2 D = 1/(cos^2 D) - (sin^2 D)/(cos^2 D) = # #= (1 - sin^2 D)/(cos ^2 D) = (cos^2 D)/(cos^2 D) = 1# 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 2010 views around the world You can reuse this answer Creative Commons License