`(sinA+2sin3A+sin5A)/(sin3A+2sin5A+sin7A)=(sin3A)/(sin5A)`

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(sin A+2sin3A+sin5A)/(sin3A+2sin5A+sin7A)=(sin3A)/(sin5A)

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

(sin A + 2 sin 3 A + sin 5 A) / (sin 3 A + 2 sin 5 A + sin 7 A) = (sin3A) /(sin 5 A)

sin A+sin3A+sin5A+sin7A=

sin A+ sin 3A+ sin 5A + sin 7A=

sin A sin2A + sin3A sin6A-sin4A sin5A = 0

( 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 )=

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

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