sin9A/sin3A-cos9A/cos3A=2
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Answer:
sin9A/sin3A - cos9A/cos3A
= (sin9A cos3A - cos9A sin3A) / sin3A cos3A
= sin(9A - 3A) / sin3A cos3A
= sin6A / sin3A cos3A
= sin6A / sin(3A + 3A) + sin(3A - 3A)/2
= 2sin6A / sin6A
= 2
hence proved
formula used:- (1) 2sinAcosB= sin(A+B) + sin(A-B)
(2) sin(A − B) = sin A cos B − cos A sin B
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