How do you prove #sinx/(1-cosx) + (1-cosx)/sinx = 2csc x#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer maganbhai P. Aug 10, 2018 Please see below. Explanation: We take , #LHS=sinx/(1-cosx)+(1-cosx)/sinx# #LHS=sin^2x/(sinx(1-cosx))+(1-cosx)/sinx# #LHS#=#(1-cos^2x)/(sinx(1-cosx))+(1-cosx)/sinxto[because sin^2x+cos^2x#=#1]# #LHS=((1-cosx)(1+cosx))/(sinx(1-cosx))+(1-cosx)/sinx# #LHS=(1+cosx)/sinx+(1-cosx)/sinx# #LHS=(1+cosx+1-cosx)/sinx# #LHS=2/sinx# #LHS=2cscx# #LHS=RHS# 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 2480 views around the world You can reuse this answer Creative Commons License