Prove that $\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}$
Prove that $\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}$
It is known that
$\cos A - \cos B = - 2\sin \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right),$
$\sin A - \sin B = 2\cos \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)$
$\therefore$ $L.H.S.$ $=\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}$
$=\frac{-2 \sin \left(\frac{9 x+5 x}{2}\right) \cdot \sin \left(\frac{9 x-5 x}{2}\right)}{2 \cos \left(\frac{17 x+3 x}{2}\right) \cdot \sin \left(\frac{17 x-3 x}{2}\right)}$
$=\frac{-2 \sin 7 x \cdot \sin 2 x}{2 \cos 10 x \cdot \sin 7 x}$
$=-\frac{\sin 2 x}{\cos 10 x}$
$=R. H.S.$
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