If ${\mathrm{k}}_1=\tan 27\mathrm{\theta }-\tan \mathrm{\theta }$ and ${\mathrm{k}}_2=\frac{\sin \mathrm{\theta }}{\cos 3\mathrm{\theta }}+\frac{\sin 3\mathrm{\theta }}{\cos 9\mathrm{\theta }}+\frac{\sin 9\mathrm{\theta }}{\cos 27\mathrm{\theta }}$ then

We can write
$\begin{array}{c}{\mathrm{k}}_1=\tan 27\mathrm{\theta }-\tan 9\mathrm{\theta }+\tan 9\mathrm{\theta }-\tan 3\mathrm{\theta }+\tan 3\mathrm{\theta }-\tan \mathrm{\theta }\\ \text{ But }\tan 3\mathrm{\theta }-\tan \mathrm{\theta }=\frac{\sin 3\mathrm{\theta cos}\mathrm{\theta }-\cos 3\mathrm{\theta sin}\mathrm{\theta }}{\cos 3\mathrm{\theta cos}\mathrm{\theta }}\\ =\frac{\sin 2\mathrm{\theta }}{\cos 3\mathrm{\theta cos}\mathrm{\theta }}\\ =\frac{2\sin \mathrm{\theta }}{\cos 3\mathrm{\theta }}\\ \therefore {\mathrm{k}}_1=2\left[\frac{\sin 9\mathrm{\theta }}{\cos 27\mathrm{\theta }}+\frac{\sin 3\mathrm{\theta }}{\cos 9\mathrm{\theta }}+\frac{\sin \mathrm{\theta }}{\cos 3\mathrm{\theta }}\right]=2{\mathrm{k}}_2\end{array}$