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 15344 views around the world You can reuse this answer Creative Commons License