Show that $\frac {\sin x} {\cos 3x} + \frac {\sin 3x} {\cos 9x} + \frac {\sin 9x} {\cos 27x} = \frac 12 (\tan 27x - \tan x)$ The question asks to prove that - 
$$\frac {\sin x} {\cos 3x} + \frac {\sin 3x} {\cos 9x} + \frac {\sin 9x} {\cos 27x}  = \frac 12 (\tan 27x - \tan x) $$
I tried combining the first two or the last two fractions on the L.H.S to allow me to use the double angle formula and get $\sin 6x$ or $\sin 18x$ but that did not help at all. 
I'm pretty sure that if I express everything in terms of $x$, the answer will ultimately appear but I'm also certain that there must be another simpler way. 

For $\cos y\ne0,$ $$\tan3y-\tan y=\dfrac{\sin(3y-y)}{\cos3y\cos y}=\dfrac{2\sin y\cos y}{\cos3y\cos y}$$
$$\implies\tan3y-\tan y=2\dfrac{\sin y}{\cos3y}$$
Set $y=x,3x,9x$  and add to recognize the Telescoping series
