`x=cosA+cos2A+cos3A , y=sinA+sin2A+sin3A=> x/y=`

(cos5A+cos3A+cosA)/(sin5A-sin3A+sinA)=

(sinA + sin3A) / (cosA + cos3A) = tanalpha, (cosA-cos2A) / (sinA + sin2A) = tanbeta, (sinA + sin3A + sin5A + sin7A) / (cosA + cos3A + cos5A + cos7A) = tan gamma , (cosA-cos5A-cos9A + cos13A) / (sinA-sin5A + sin9A-sin13A) = tandelta, then (a) alpha = 4beta (b) gamma = 2alpha (c) gamma = delta (d) none

(cos3A+sin3A)/(cosA-sinA)=

Prove that ((cos^3A-sin^3A)/(cosA - sinA))- ((cos^3A + sin^3A)/(cosA + sinA))= 2sin A cos A

If 2cos x+2cos3x=cos y and 2sin x+2sin3x=sin y then cos2x=

If 2cos x+2cos3x=cos y and 2sin x+2sin3x=sin y then cos2x=

If 2cos x+2cos3x=cos y and 2sin x+2sin3x=sin y then cos2x=

Prove that: (sin5A-sin3A)/(cos5A+cos3A)=tanA (2) (sinA+sin3A)/(cosA+cos3A)=tan2A

Prove that: (sin5A-sin3A)/(cos5A+cos3A)=tanA (sinA+sin3A)/(cosA+cos3A)=tan2A