Question

Trigonometric series of the form
$\frac{\sin (A-B)}{\cos A\cdot \cos B}+\frac{\sin (B-C)}{\cos B\cdot \cos C}+\frac{\sin (C-D)}{\cos C\cdot \cos D}$
$=\tan A-\tan D$
As we know that,
$\frac{\sin (A-B)}{\cos A\cdot \cos B}=\tan A-\tan B$
Based on the above given information, find sum of the series
$\frac{\sin x}{\cos 3x}+\frac{\sin 3x}{\cos 9x}+\frac{\sin 9x}{\cos 27x}+\cdots$ upto $n$ terms

• A. $\frac{1}{2}(\cot 3^{n}x-\cot x)$
• B. $\frac{1}{2}(\tan 3^{n}x-\tan x)$
• C. $\frac{1}{2}(\tan 3x-\tan 3^{n-1}x)$
• D. $\frac{1}{2}(\cot 3x-\cot 3^{n-1}x)$

Answer 1

The correct option is B $\frac{1}{2}(\tan 3^{n}x-\tan x)$

$S_n=\frac{\sin x}{\cos 3x}+\frac{\sin 3x}{\cos 9x}+\frac{\sin 9x}{\cos 27x}+\cdots +\frac{\sin \left(3^{n-1}\right)x}{\cos \left(3^n\right)x}$
$=\frac{1}{2}[\frac{2\sin x\cos x}{\cos 3x\cos x}+\frac{2\sin 3x\cos 3x}{\cos 9x\cos 3x}\cdots +\frac{2\sin \left(3^{n-1}\right)x\cos \left(3^{n-1}\right)x}{\cos \left(3^n\right)x\cos \left(3^{n-1}\right)x}]$
$=\frac{1}{2}[\frac{\sin 2x}{\cos 3x\cos x}+\frac{\sin 6x}{\cos 9x\cos 3x}\cdots +\frac{\sin 2\cdot \left(3^{n-1}\right)x}{\cos \left(3^n\right)x\cos \left(3^{n-1}\right)x}]$
$=\frac{1}{2}[\frac{\sin (3x-x)}{\cos 3x\cos x}+\frac{\sin (9x-3x)x}{\cos 9x\cos 3x}\cdots +\frac{\sin (3^{n}x-3^{n-1})x}{\cos \left(3^n\right)x\cos \left(3^{n-1}\right)x}]$
$=\frac{1}{2}\left[\right(\tan 3x-\tan x)+(\tan 9x-\tan 3x)+\cdots +(\tan \left(3^n\right)x-\tan \left(3^{n-1}\right)x\left)\right]$
$=\frac{1}{2}\left(\tan \right(3^n)x-\tan x)$

Alternate solution:
Substitute $n=2$, $x=0$ and $x=\frac{\pi }{4}$ in the options
and then verify it with the given series