How can one prove that #ln(sec^2x+csc^2x) = ln(sec^2x)+ln(csc^2 x)#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Shwetank Mauria Feb 15, 2017 Please see below. Explanation: #sec^2x+csc^2x# = #1/cos^2x+1/sin^2x# = #(sin^2x+cos^2x)/(sin^2xcos^2x)# = #1/(sin^2xcos^2x)# = #sec^2xcsc^2x# As #sec^2x+csc^2x=sec^2xcsc^2x#, taking natural log on both sides #ln(sec^2x+csc^2x)=lnsec^2x+lncsc^2x# 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 1322 views around the world You can reuse this answer Creative Commons License