<s>[INST] 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. 
 [/INST] 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 $\tan\left(3^{n}x\right)$.
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