`sinA+sin3A+...+sin(2n-1)A=(sin^2n A)/(sinA)`

prove that : (sinA+2sin3A+sin5A)/(sin3A+2sin5A+sin7A)=(sin3A)/(sin5A)

Prove that : cos A cos 2A cos 2^2 A cos 2^3 A........ cos 2^(n-1) A= (sin 2^n A)/(2^n sinA) .

If sin2A=2sinA then A=

(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

4sinA.sin(60^(@)+A)sin(60^(@)-A)=sin3A

4sinA.sin(60^(@)+A)sin(60^(@)-A)=sin3A

Prove that: sinA. Sin(60^(2)+A).sin(120^(@)+A)=1/4sin3A

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

sqrt((1+sinA)/(1-sin A ))=