$\text{Show that }\frac{\cos 2\alpha +\cos 5\alpha +\cos 8\alpha }{\sin 2\alpha +\text{sin}5\alpha +\sin 8\alpha }=\cot 5\alpha $
$\begin{align} & \frac{\cos 2\alpha +\cos 5\alpha +\cos 8\alpha }{\sin 2\alpha +\sin 5\alpha +\sin 8\alpha }=\frac{\cos 2\alpha +\cos 8\alpha +\cos 5\alpha }{\sin 2\alpha +\sin 8\alpha +\sin 5\alpha } \\ & \frac{\cos 2\alpha +\cos 5\alpha +\cos 8\alpha }{\sin 2\alpha +\sin 5\alpha +\sin 8\alpha }=\frac{2\cos 5\alpha \cos 3\alpha +\cos 5\alpha }{2\sin 5\alpha \cos 3\alpha +\sin 5\alpha } \\ & \frac{\cos 2\alpha +\cos 5\alpha +\cos 8\alpha }{\sin 2\alpha +\sin 5\alpha +\sin 8\alpha }=\frac{\cos 5\alpha (2\cos 3\alpha +1)}{\sin 5\alpha (2\cos 3\alpha +1)}=\frac{\cos 5\alpha }{\sin 5\alpha } \\ & \frac{\cos 2\alpha +\cos 5\alpha +\cos 8\alpha }{\sin 2\alpha +\sin 5\alpha +\sin 8\alpha }=\cot 5\alpha \\\end{align}$