Question

Show that $\frac{\sin x}{\cos 3x}+\frac{\sin 3x}{\cos 9x}+\frac{\sin 9x}{\cos 27x}=\frac{1}{2}[\tan 27x-\tan x]$.

Answer 1

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

multiplying and dividing by 2
$=\frac{1}{2}[\frac{2\sin x}{\cos 3x}+\frac{2\sin 3x}{\cos 9x}+\frac{2\sin 9x}{\cos 27x}]$
multiplying and dividing by $cosx,cos9x,cos27$ respectively and simplifying
$=\frac{1}{2}[\frac{\sin 2x}{\cos 3x\cos x}+\frac{\sin 6x}{\cos 9x\cos 3x}+\frac{\sin 18x}{\cos 27xcos9x}]$

Use, $\sin (A-B)=\sin A\cos B-\cos A\sin B$

$=\frac{1}{2}\left[\frac{\sin 3x\cos x-\cos 3x\sin x}{\cos 3x\cos x}\right]+\frac{1}{2}\left[\frac{\sin 9x\cos 3x-\cos 9x\sin 3x}{\cos 27x\cos 9x}\right]$
$+\frac{1}{2}\left[\frac{\sin 27x\cos 9x-\cos 27x\sin 9x}{\cos 27x\cos 9x}\right]$
$=\frac{1}{2}[\tan 27x-\tan 9x+\tan 9x-\tan 3x+\tan 3x-\tan x]=\frac{1}{2}[\tan 27x-\tan x]$