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The period of $\frac{\sin x}{\cos 3 x}+\frac{\sin 3 x}{\cos 9 x}+\frac{\sin 9 x}{\cos 27 x}+\frac{\sin 27 x}{\cos 81 x}$ is
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The correct answer is:
$\pi$
Given, $\frac{\sin x}{\cos 3 x}+\frac{\sin 3 x}{\cos 9 x}+\frac{\sin 9 x}{\cos 27 x}+\frac{\sin 27 x}{\cos 81 x}$
$=\frac{1}{2}\left[\frac{2 \sin x \cos x}{\cos 3 x \cos x}+\frac{2 \sin 3 x \cos 3 x}{\cos 9 x \cos 3 x}+\frac{2 \sin 9 x \cos 9 x}{\cos 27 x \cos 9 x}\right.$ $\left.+\frac{2 \sin 27 x \cos 27 x}{\cos 81 x \cos 27 x}\right]$
$=\frac{1}{2}\left[\frac{\sin 2 x}{\cos 3 x \cos x}+\frac{\sin 6 x}{\cos 9 x \cos 3 x}+\frac{\sin 18 x}{\cos 27 x \cos 9 x}\right.$ $\left.+\frac{\sin 54 x}{\cos 81 x \cos 28 x}\right]$
$=\frac{1}{2}\left[\frac{\sin (3 x-x)}{\cos 3 x \cos x}+\frac{\sin (9 x-3 x)}{\cos 9 x \cos 3 x}+\frac{\sin (27 x-9 x)}{\cos 27 x \cos 9 x}\right.$ $\left.+\frac{\sin (81 x-27 x)}{\cos 81 x \cos 27 x}\right]$
$=\frac{1}{2}[\tan 3 x-\tan x+\tan 9 x-\tan 3 x+\tan 27 x$ $-\tan 9 x+\tan 81 x-\tan 27 x]$
$=\frac{1}{2}[\tan (81 x)-\tan x]$
$\therefore$ Required period is $\pi$.
$=\frac{1}{2}\left[\frac{2 \sin x \cos x}{\cos 3 x \cos x}+\frac{2 \sin 3 x \cos 3 x}{\cos 9 x \cos 3 x}+\frac{2 \sin 9 x \cos 9 x}{\cos 27 x \cos 9 x}\right.$ $\left.+\frac{2 \sin 27 x \cos 27 x}{\cos 81 x \cos 27 x}\right]$
$=\frac{1}{2}\left[\frac{\sin 2 x}{\cos 3 x \cos x}+\frac{\sin 6 x}{\cos 9 x \cos 3 x}+\frac{\sin 18 x}{\cos 27 x \cos 9 x}\right.$ $\left.+\frac{\sin 54 x}{\cos 81 x \cos 28 x}\right]$
$=\frac{1}{2}\left[\frac{\sin (3 x-x)}{\cos 3 x \cos x}+\frac{\sin (9 x-3 x)}{\cos 9 x \cos 3 x}+\frac{\sin (27 x-9 x)}{\cos 27 x \cos 9 x}\right.$ $\left.+\frac{\sin (81 x-27 x)}{\cos 81 x \cos 27 x}\right]$
$=\frac{1}{2}[\tan 3 x-\tan x+\tan 9 x-\tan 3 x+\tan 27 x$ $-\tan 9 x+\tan 81 x-\tan 27 x]$
$=\frac{1}{2}[\tan (81 x)-\tan x]$
$\therefore$ Required period is $\pi$.
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