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Prove that: $$\frac {2\sin x}{\cos 3x}+\frac {2\sin 3x}{\cos 9x}+\frac {2\sin 9x}{\cos 27x}=\tan 27x - \tan x$$

My Work,

If we multiply the numerator and denominator of first, second and third term of left hand side by $\cos x$, $\cos 3x$ and $\cos 9x$ respectively and further use the compound angle formula, then we get RHS. But, I want to know if there is any other alternative solution to this problem??

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    Note that $\sin x$ is implemented in $\LaTeX$ as `\sin x`. – 2017-02-07
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    See also : http://math.stackexchange.com/questions/1299957/show-that-frac-sin-x-cos-3x-frac-sin-3x-cos-9x-frac-sin-9x – 2017-02-07

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