`(sina*sin3a+sin3a .sin7a+sin5a.sin15a)/(sina*cos3a+sin3a cos7a+sin5a.cos15a)=`

(sin alpha*sin3 alpha+sin3 alpha*sin7 alpha+sin5 alpha*sin15 alpha)/(sin alpha*cos3 alpha+sin3 alpha cos7 alpha+sin5 alpha*cos15 alpha)=

(sin A sin2A + sin3A sin6A) / (sin A cos2A + sin3A cos6A) = tan5A

(sin3A)/(sinA)-(cos3A)/(cosA)=2

( sin alpha sin 3alpha + sin 3alpha sin 7 alpha + sin 5 alpha sin 15 alpha )/( sin alpha cos 3 alpha + sin 3alpha cos 7 alpha + sin 5 alpha cos 15 alpha ) =

Prove that: (sin2A sin3A-sin4A sin5A+sin3A sin6A)/(cos2A cos3A-cos4A cos5A+cos3A cos6A)=tan A tan4A

(sin A+sin3A+sin5A+sin7A)/(cos A+cos3A+cos5A+cos7A)=tan4A

( sin 4 A sin 3A - sin 5A sin 2A+ sin 7A sin 4A )/(cos 2A cos 3A - cos 2A * cos 7A + cos A * cos 10A )=

(sinA+sin3A+sin5A+sin7A)/(cosA+cos3A+cos5A+cos7A) is :

(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

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