How do you prove #(2-sec^2[theta])/(sec^2[theta]) =1-2sin^2[theta]#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Cesareo R. Oct 7, 2016 see below. Explanation: #(2-sec^2 theta)/(sec^2 theta ) =1-2sin^2 theta# #(2-sec^2 theta )/(sec^2 theta ) = (2 cos^2theta-1)/cos^2theta cos^2theta = 2cos^2theta-1=2(1-sin^2theta)-1 = 1-2sin^2theta# 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 3188 views around the world You can reuse this answer Creative Commons License