Q60 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Mathematics
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
- A$\frac{2 \pi}{3}$
- B$\frac{\pi}{81}$
- C$2 \pi$
- Correct answerD$\pi$
Solution
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 $\pi$.
Hence, the correct answer is option: 'D'.