<s>[INST] 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
$$\frac{\sin x}{\cos 3x}+\frac{\sin 3x}{\cos 9x}+\frac{\sin 9x}{\cos 27x} = \frac{1}{2}\left(\tan 27x-\tan x\right)$$
 [/INST] 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 $9x$
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