How do you verify #(1-cos2t)/ sin t= 2 sin t#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer GiĆ³ May 13, 2015 use the fact that #cos2t=cos^2t-sin^2t# to get #(1-cos^2t+sin^2t)/sint=2sint# But #1=sin^2t+cos^2t# So: #(sin^2t+cancel(cos^2t)-cancel(cos^2t)+sin^2t)/sint=2sint# #(2sin^cancel(2)t)/cancel(sint)=2sint# #2sint=2sint# 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 2590 views around the world You can reuse this answer Creative Commons License