Question

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

• A. $\frac{1}{2}(\cot x-\cot 81x)$
• B. $\frac{1}{2}(\cot 81x-\cot x)$
• C. $(\cot x-\cot 81x)$
• D. $(\cot x+\cot 81x)$

Answer 1

The correct option is A $\frac{1}{2}(\cot x-\cot 81x)$

$t_1=\frac{\cos x}{\sin 3x}=\frac{1}{2}\frac{2\sin x\cos x}{\sin x\sin 3x}$
$=\frac{1}{2}\frac{\sin 2x}{\sin x\sin 3x}$
$=\frac{1}{2}\frac{\sin (3x-x)}{\sin x\sin 3x}$
$=\frac{1}{2}\frac{\sin 3x\cos x-\cos 3x\sin x}{\sin x\sin 3x}$
$=\frac{1}{2}(\cot x-\cot 3x)$
Writing $t_2,t_3,t_4$ in a similar way,
$t_1+t_2+t_3+t_4=\frac{1}{2}(\cot x-\cot 81x)$