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TS EAMCET 11-Sep-2020 Shift 1 Solved Paper

Section: Mathematics

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Question : 20 of 160

Marks: +1, -0

\frac{\sin x}{\cos 3x} + \frac{\sin 3x}{\cos 9x} + \frac{\sin 9x}{\cos 27x} + \frac{\sin 27x}{\cos 81x}

$\frac{2\pi}{3}$
$\frac{\pi}{81}$
$2\pi$
$\pi$

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} + \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} + \frac{\sin 54 x}{\cos 81 \times \cos 28 x} \right]

= \frac{1}{2} \left[ \frac{\sin (3x-x)}{\cos 3x \cos x} + \frac{\sin (9x-3x)}{\cos 9x \cos 3x} + \frac{\sin (27x-9x)}{\cos 27x \cos 9x} + \frac{\sin (81x-27x)}{\cos 81x \cos 27x} \right]

= \frac{1}{2} \left[ \tan 3 x \;\; -\tan x+\tan 9 x-\tan 3 x+\tan 27 x-\tan 9 x+\tan 81 x-\tan 27 x \right]

= \frac{1}{2} \left[ \tan (81x) - \tan x \right]

\therefore

Required period is

\pi .

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