Answer
$\tan 2 \ \theta$
Work Step by Step
Recall the Sum to Product Identities:
$a) \sin x +\sin y =2 \sin \dfrac{x+y}{2} \cos \dfrac{x-y}{2} \\ b) \sin x - \sin y =2 \sin \dfrac{x -y}{2} \cos \dfrac{x +y}{2} \\ c) \cos x +\cos y =2 \cos \dfrac{x+y}{2} \cos \dfrac{x-y}{2} \\ d) \cos x - \cos y = -2 \sin \dfrac{x+y}{2} \sin \dfrac{x-y}{2}$
By identities $(a)$ and (c), we have:
$\dfrac{\cos \theta - \cos 3 \theta}{\sin 3 \theta - \sin \theta }=\dfrac{- 2 \sin \dfrac{ \theta + 3 \theta}{2} \ \sin \dfrac{ \theta - 3 \theta }{2} }{2 \sin \dfrac{ 3 \theta - \theta}{2} \ \cos \dfrac{3 \theta + \theta }{2} }\\=\dfrac{2 \sin 2 \theta \ \sin \theta }{2 \cos 2 \theta \ \sin \theta } \\=\tan 2 \ \theta$
