How do you prove # ( 1 / (secx - tanx) ) - ( 1 / (secx + tanx ) ) = 2 tan x#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Arunraju Naspuri Jul 4, 2015 It can be proved by #sec^2x-tan^2x=1# Explanation: #sec^2x-tan^2x=1# #(secx+tanx)(secx-tanx)=1#[since#a^2-b^2=(a+b)(a-b)]# #secx+tanx=1/(secx-tanx)##" " color(red)((1))# #secx-tanx=1/(secx+tanx)##" " color(red)((2))# #LHS= (1/(secx−tanx)−1/(secx+tanx))# #" " color(red)((1))#&#color(red)((2))# substitute in the above equation #LHS=(secx+tanx)-(secx-tanx)# #LHS=secx+tanx-secx+tanx# #LHS=2tanx# #LHS=RHS# Hence proved 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 5857 views around the world You can reuse this answer Creative Commons License