How do I prove #(sec x/cos x)+(csc x/sin x)=4 csc^2(2x)#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Abhishek K. Nov 21, 2017 Please refer to the explanation below for proof . Explanation: #LHS=secx/cosx+cscx/sinx# #=(1/cosx)/cosx+(1/sinx)/sinx# #=1/cos^2x+1/sin^2x# #=(sin^2x+cos^2x)/(sin^2x*cos^2x# #=1/(sin^2x*cos^2x)*4/4# #=4/(2sinx*cosx)^2# #=4/sin^2(2x)=4csc^2(2x)=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 1618 views around the world You can reuse this answer Creative Commons License