Question

(sinasin2a+sin3asin6a+sin4asin13a)/(sinacos2a+sin3asin6a+sin4acos13a)= tan9a

Answer 1

post it through equation forum please

Answer 2

\[sinAsin2A = \frac{1}{2}(\cos A - \cos 3A)\]
\[\sin3Asin6A = \frac{1}{2}(\cos3A - \cos9A)\]
\[\sin4Asin13A = \frac{1}{2}(\cos9A - \cos17A)\]

\[sinAcos2A = \frac{1}{2}(-sinA+\sin3A) \]\[\sin3Acos6A = \frac{1}{2}(-\sin3A+\sin9A)\]
\[\sin4Acos13A = \frac{1}{2}(-\sin9A+\sin17A)\]

\[(sinAsin2A + \sin3Asin6A + \sin4Asin13A) / (sinAcos2A + \sin3Acos6A + \sin4Acos13A) * 2 / 2\]
\[= (\cos A - \cos 3A + \cos3A - \cos9A + \cos9A - \cos17A) / (-sinA+\sin3A -\sin3A+\sin9A -\sin9A+\sin17A)\]
\[= (\cos A - \cos 17A) / (\sin 17A-\sin A)\]
\[= 2 \sin 9A \sin 8A / 2 \cos 9A \sin 8A\] (cancel 2 and sin 8A)
\[= \sin 9A / \cos 9A\]
\[= \tan 9A\]