Prove that ??#(Sinx+Sin2x+Sin3x)/(cosx+cos2x+cos3x) = tan2x# Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Abhishek K. May 11, 2018 #LHS=(sinx+sin2x+sin3x)/(cosx+cos2x+cos3x)# #=(2sin((3x+x)/2)*cos((3x-x)/2)+sin2x)/(2cos((3x+x)/2)*cos((3x-x)/2)+cos2x# #=(2sin2x*cosx+sin2x)/(2cos2x*cosx+cos2x)# #=(sin2xcancel((1+2cosx)))/(cos2xcancel((1+2cosx)))# #=tan2x=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 14822 views around the world You can reuse this answer Creative Commons License