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The value of the expression

$$\dfrac{\sin x}{ \cos 3x} + \dfrac{\sin 3x}{ \cos 9x} + \dfrac{\sin 9x}{ \cos 27x}$$ in terms of $\tan x$ is

My Approach

If I take L.C.M of this as $\cos 3 \cos 9x \cos 27x$ and respectively multiply the numerator then it is getting very lengthy. Even if I will use the identity of $\sin 3x$ then also I am not getting appropriate answer. Please suggest some nice and short way of doing this question.

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marked as duplicate by lab bhattacharjee trigonometry Jul 10 '16 at 15:50

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    $\begingroup$ Instead of trying to combine the three terms (your "LCM" approach), I would just do the first term. The other terms can then be expressed in terms of substituting $3x$ and $9x$ for $x$, respectively. $\endgroup$ – hardmath Jul 9 '16 at 18:42
  • $\begingroup$ Hint: Show that sinx /cosx = (1/2)(tan3x - tan x) $\endgroup$ – GohP.iHan Jul 10 '16 at 9:09
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$$ {\sin x \over \cos 3x}$$ $$= {\sin x \over \cos x(1 - 4\sin^2 x)}$$ $$= {\tan x \over (1 - 4\sin^2 x)}$$

Now you can get $$\sin^2 x = {\tan^2 x \over 1 + \tan^2 x}$$ from $$ \cot^2 x + 1= { 1\over \sin^2 x}$$

$$\therefore {\sin x \over \cos 3x} = {\tan x \over (1 - 4\sin^2 x)} = {\tan x (1 + \tan^2 x) \over (1 - 3\tan^2 x)}$$

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