Prove that


Prove that

$\frac{\cos 9 x-\cos 5 x}{\cos 17 x-\sin 3 x}=\frac{-\sin 2 x}{\cos 10 x}$



$=\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}$

$=\frac{-2 \sin \frac{9 x+5 x}{2} \sin \frac{9 x-5 x}{2}}{2 \cos \frac{17 x+3 x}{2} \sin \frac{17 x-3 x}{2}}$

$=\frac{-2 \sin 7 x \sin 2 x}{2 \cos 10 x \sin 7 x}$

$=\frac{-\sin 2 x}{\cos 10 x}$

Using the formula,

$\cos A-\cos B=-2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$

$\sin A-\sin B=2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$


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