If $k_1=\tan 27\theta -\tan \theta$  and $k_2=\frac{\sin \theta }{\cos 3\theta }+\frac{\sin 3\theta }{\cos 9\theta }+\frac{\sin 9\theta }{\cos 27\theta }$then  

We can write

 $k_1=\tan 27\theta -\tan 9\theta +\tan 9\theta -\tan 3\theta +\tan 3\theta -$$\tan \theta$

But $\begin{array}{r}\tan 3\theta \tan \theta -\frac{\sin 3\theta \cos \theta -\cos 3\theta \sin \theta }{\cos 3\theta \cos \theta }\\ =\frac{\sin 2\theta }{\cos 3\theta \cos \theta }=\frac{2\sin \theta }{\cos 3\theta }\\ k_1=2\left[\frac{\sin 9\theta }{\cos 27\theta }+\frac{\sin 3\theta }{\cos 9\theta }+\frac{\sin \theta }{\cos 3\theta }\right]=2k_2.\end{array}$