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??


marked as duplicate by S.C.B., Nosrati, lab bhattacharjee trigonometry Feb 7 '17 at 8:21

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