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. 
 A: In general, we have
$$\sum_{k=1}^n \dfrac{\sin\left(3^{k-1}x \right)}{\cos\left(3^kx\right)} = \dfrac{\tan\left(3^nx\right)-\tan(x)}2$$
We can prove this using induction. $n=1$ is trivial. All we need to make use of is that
$$\tan(3t) = \dfrac{3\tan(t)-\tan^3(t)}{1-3\tan^2(t)}$$
For the inductive step, we have
$$\sum_{k=1}^n \dfrac{\sin\left(3^{k-1}x \right)}{\cos\left(3^kx\right)} = \dfrac{\tan\left(3^nx\right)-\tan(x)}2$$
Adding the last term, we obtain
$$\sum_{k=1}^{n+1} \dfrac{\sin\left(3^{k-1}x \right)}{\cos\left(3^kx\right)} = \dfrac{\tan\left(3^nx\right)-\tan(x)}2 + \dfrac{\sin\left(3^{n}x \right)}{\cos\left(3^{n+1}x\right)} = \dfrac{\tan\left(3^{n+1}x\right)-\tan(x)}2$$
where we again rely on $(\spadesuit)$ to express $\tan\left(3^{n+1}x\right)$ in terms of $\tan\left(3^{n}x\right)$.
A: @MayankJain @user,
I don't know that the case for $n=1$ is trivial, especially for someone in a trig class currently. I will offer a proof without induction using substitution instead. Mayank, to see that this is the case for $n=1$, convert the RHS of the equation as follows:
$\dfrac{1}{2} \cdot \big{[}\dfrac{\sin 3x}{\cos 3x} - \dfrac{\sin x}{\cos x}]$ = (1) $\dfrac{1}{2} \cdot \big(\dfrac{\sin 3x \cos x - \cos 3x \sin x}{\cos 3x \cos x}\big{)} $ = (2) $ \dfrac{1}{2} \cdot \dfrac{\sin 2x}{\cos 3x \cos x}$ = (3) $\dfrac{1}{2} \cdot \dfrac{2\sin x \cos x}{\cos 3x \cos x}$ = $\dfrac{\sin x}{\cos 3x}$
This is pretty heavy on trig identities. We get equivalence (1) by multiplying out the fraction, equivalence (2) because $\sin(u - v) = \sin u \cos v - \cos u \sin v$, equivalence (3) because $\sin 2x = 2\sin x \cos x$, and, finally, (4) by cancellation. 
Now, if you are unfamiliar with induction, substitution will help here as an alternative method. Since you can now derive the equivalence for the 'trivial' case, set $3x = u$ for the second case, and $9x = v$ for the third. Then you already know $\dfrac{\sin u}{\cos 3u} = \dfrac{1}{2} \cdot (\tan 3u - \tan u)$, and similarly, $\dfrac{\sin v}{\cos 3v} = \dfrac{1}{2} \cdot (\tan 3v - \tan v)$. So now we can re-write the original expression $\dfrac{\sin x}{\cos 3x} + \dfrac{\sin u}{\cos 3u} + \dfrac{\sin v}{\cos 3v}$ as $\dfrac{1}{2} \cdot (\tan(3x) - \tan x + \tan(9x) - \tan(3x) + \tan(27x) - \tan(9x))$ and everything cancels out except the desired expression: $\dfrac{1}{2} \cdot (\tan(27x) - \tan(x))$.
A: 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
