How do you prove #(sinx-2sin2x+sin3x)/(sinx+2 sin2x+sin3x)= -tan^2(x/2)#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer P dilip_k Jun 8, 2018 #LHS=(sinx-2sin2x+sin3x)/(sinx+2 sin2x+sin3x)# #=(2sin2xcosx-2sin2x)/(2sin2xcosx+2 sin2x)# #=-(2sin2x(1-cosx))/(2sin2x(1+cosx))# #=-(2sin^2(x/2))/(2cos^2(x/2))# #= -tan^2(x/2)=RHS# 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 3026 views around the world You can reuse this answer Creative Commons License