If the expression $\frac{\sin \theta \sin 2\theta +\sin 3\theta \sin 6\theta +\sin 4\theta \sin 13\theta }{\sin \theta \cos 2\theta +\sin 3\theta \cos 6\theta +\sin 4\theta \cos 13\theta }=\tan k\theta$, where $k\in N$ then the value of $k=$

$\begin{array}{l}\frac{\sin \theta \sin 2\theta +\sin 3\theta \sin 6\theta +\sin 4\theta \sin 13\theta }{\sin \theta \cos 2\theta +\sin 3\theta \cos 6\theta +\sin 4\theta \cos 13\theta }\\ =\frac{\cos \theta -\cos 3\theta +\cos 3\theta -\cos 9\theta +\cos 9\theta -\cos 17\theta }{\sin 3\theta -\sin \theta +\sin 9\theta -\sin 3\theta +\sin 17\theta -\sin 9\theta }\\ =\frac{\cos \theta -\cos 17\theta }{\sin 17\theta -\sin \theta }=\frac{2\sin 9\theta \sin 8\theta }{2\sin 8\theta \cos 9\theta }=\tan 9\theta \\ \therefore k=9\end{array}$