Question #4d95b Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Alyssa Apr 8, 2017 #(sec x + csc x)/(sin x + cosx) * sin 2x = 2# Taking the left hand side, #(sec x + csc x)/(sin x + cosx) * sin 2x# = #(1/cosx + 1/sin x)/(sin x + cos x) * sin 2x# = #((sin x + cos x)/(sin x cos x))/(sinx + cosx) * sin 2x# #sin x + cos x# cancel out recall the identity #sin 2x = 2 sin x cos x# = #1/(sin x cos x) * 2 sin x cos x# = #2# 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 1211 views around the world You can reuse this answer Creative Commons License