`(sinAsin2A+sin3Asin6A+sin4Asin13A)/(sinAcos2a+sin3Acos6A+sin4Acos13A)`=`tan9A`

(sin A sin2A + sin3A sin6A + sin4A sin13A) / (sin A cos2A + sin3A cos6A + sin4A cos13A) = tan9A

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

2sinAcos^3A-2sin^3AcosA=

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

(sin A-sin5A+sin9A-sin13A)/(cos A-cos5A-cos9A+cos13A)=cot4A

(sin A-sin5A+sin9A-sin13A)/(cos A-cos5A-cos9A+cos13A)=cot4A

sin2A+2sin4A+sin6A=4cos^2Asin4A

(sin6A+sin2A+2sin4A)/(sin7A+sin3A+2sin5A)=(sin4A)/(sin5A)

(sin6A+sin2A+2sin4A)/(sin7A+sin3A+2sin5A)=(sin4A)/(sin5A)

Prove that: (sin3A cos4A-sin Acos2A)/(sin4A sin A+cos6A cos A)=tan2A