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Prove that $\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}$
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$\begin{aligned} \text { LHS } &=\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x} \\ &=\frac{-2 \sin \frac{9 x+5 x}{2} \sin \frac{9 x-5 x}{2}}{2 \cos \frac{17 x+3 x}{2} \sin \frac{17 x-3 x}{2}} \end{aligned}$
$\left[\begin{array}{l}\because \cos \theta-\cos \phi=2 \sin \frac{\theta+\phi}{2} \sin \frac{\theta-\phi}{2} \\ \sin \theta-\sin \phi=2 \cos \frac{\theta+\phi}{2} \sin \frac{\theta-\phi}{2}\end{array}\right]$
$=\frac{-\sin 7 x \cdot \sin 2 x}{\cos 10 x \cdot \sin 7 x}=-\frac{\sin 2 x}{\cos 10 x}=\mathrm{RHS}$
$\left[\begin{array}{l}\because \cos \theta-\cos \phi=2 \sin \frac{\theta+\phi}{2} \sin \frac{\theta-\phi}{2} \\ \sin \theta-\sin \phi=2 \cos \frac{\theta+\phi}{2} \sin \frac{\theta-\phi}{2}\end{array}\right]$
$=\frac{-\sin 7 x \cdot \sin 2 x}{\cos 10 x \cdot \sin 7 x}=-\frac{\sin 2 x}{\cos 10 x}=\mathrm{RHS}$
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