Show that $\frac{\sin x}{\cos 3x}+\frac{\sin 3x}{\cos 9x}+\frac{\sin 9x}{\cos 27x} = \frac{1}{2}\left(\tan 27x-\tan x\right)$

Show that

$$\frac{\sin x}{\cos 3x}+\frac{\sin 3x}{\cos 9x}+\frac{\sin 9x}{\cos 27x} = \frac{1}{2}\left(\tan 27x-\tan x\right)$$

A more general result is in fact true. $$\sum_{k=0}^{n} \dfrac{\sin\left(3^k x\right)}{\cos\left(3^{k+1} x\right)} = \dfrac{\tan\left(3^{n+1} x\right) - \tan(x)}2 \,\,\,\, (\spadesuit)$$ Take $n=2$ in $(\spadesuit)$, to get what you want.

To prove $(\spadesuit)$, first note that $$\dfrac{\sin \left(3^k \cdot x \right)}{\cos \left(3^{k+1} \cdot x \right)} = \dfrac{\tan \left( 3^{k+1} \cdot x\right) - \tan \left(3^{k} \cdot x\right)}2 \,\,\,\, (\clubsuit)$$

Now telescopic summation gives you $(\spadesuit)$.

$(\clubsuit)$ is proved by noting the fact that \begin{align} \tan \left( 3^{k+1} \cdot x\right) - \tan \left(3^{k} \cdot x\right) & = \dfrac{\sin \left( 3^{k+1} \cdot x\right)}{\cos \left( 3^{k+1} \cdot x\right)} - \dfrac{\sin \left( 3^{k} \cdot x\right)}{\cos \left( 3^{k} \cdot x\right)}\\ & = \dfrac{\sin \left( 3^{k+1} \cdot x\right)\cos \left( 3^{k} \cdot x\right) - \sin \left( 3^{k} \cdot x\right)\cos \left( 3^{k+1} \cdot x\right)}{\cos \left( 3^{k+1} \cdot x\right)\cos \left( 3^{k} \cdot x\right)}\\ & = \dfrac{\sin \left(2 \cdot 3^k \cdot x\right)}{\cos \left( 3^{k+1} \cdot x\right)\cos \left( 3^{k} \cdot x\right)} = \dfrac{2 \sin \left(3^k \cdot x\right) \cos \left(3^k \cdot x\right)}{\cos \left( 3^{k+1} \cdot x\right)\cos \left( 3^{k} \cdot x\right)}\\ & = 2 \cdot \dfrac{\sin \left(3^k \cdot x \right)}{\cos \left(3^{k+1} \cdot x \right)} \end{align}

Hint:: $$\frac{\sin x}{\cos 3x}=$$ $$= \frac{1}{2} \cdot \frac{2 \cdot \sin x \cdot \cos x}{\cos x \cdot \cos 3x}$$ $$= \frac{1}{2} \cdot \frac{\sin 2x}{\cos x \cdot \cos 3x}$$ $$= \frac{1}{2} \cdot \frac{ \sin(3x - x) }{ \cos x \cdot \cos3x }$$ $$= \frac{1}{2} \cdot \frac{ \sin 3x \cdot \cos x - \cos 3x \cdot \sin x}{ \cos x \cdot \cos3x}$$ $$= \frac{1}{2} \cdot \frac {\sin 3x \cdot \cos x}{\cos x \cdot \cos 3x} - \frac{1}{2} \cdot \frac{\cos 3x \cdot \sin x}{\cos x \cdot \cos 3x}$$ $$= \frac{1}{2} \cdot \left (\frac{\sin 3x}{\cos 3x} - \frac{\sin x}{\cos x} \right )$$ $$= \frac{1}{2} \cdot \left (\tan 3x - \tan x \right )$$

Now successively replace $x$ with $3x$ and $9x$