How do you prove #csc^2 θ+ sec^2 θ = sec^2 θ csc^2 θ#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Bdub May 5, 2016 see below Explanation: Left Side: #=csc^2theta+sec^2 theta# #=1/sin^2 theta + 1/cos^2 theta# #=(cos^2 theta + sin^2 theta)/(sin^2 theta cos^2 theta)# #=1/(cos^2 theta sin^2 theta)# #=1/cos^2 theta * 1/sin^2 theta# #=sec^2 theta csc^2 theta# #=#Right Side 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 14070 views around the world You can reuse this answer Creative Commons License