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 $n^{\text{th}}$ term of the series
$\frac{\sin x}{\cos 3x}+\frac{\sin 3x}{\cos 9x}+\frac{\sin 9x}{\cos 27x}+\cdots$ upto $n$ terms

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

Answer 1

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

$t_n=\frac{\sin \left(3^{n-1}\right)x}{\cos \left(3^n\right)x}$
$=\frac{1}{2}\left(\frac{2\sin \left(3^{n-1}\right)x\cos \left(3^{n-1}\right)x}{\cos \left(3^n\right)x\cos \left(3^{n-1}\right)x}\right)$
$=\frac{1}{2}\left(\frac{\sin 2\cdot \left(3^{n-1}\right)x}{\cos \left(3^n\right)x\cos \left(3^{n-1}\right)x}\right)$
$=\frac{1}{2}\left(\frac{\sin (3^{n}x-3^{n-1}x)}{\cos \left(3^n\right)x\cos \left(3^{n-1}\right)x}\right)$
$=\frac{1}{2}\left(\tan \right(3^n)x-\tan (3^{n-1}\left)x\right)$