Question

If $p=\tan 27\theta -\tan \theta ;q=\frac{\sin \theta }{\cos 3\theta }+\frac{\sin 3\theta }{\cos 9\theta }+\frac{\sin 9\theta }{\cos 27\theta }$, then prove that $\frac{p}{q}=2$

Answer 1

Write $p=[\tan 27\theta -\tan 9\theta +\tan 9\theta -\tan 3\theta +\tan 3\theta -\tan \theta ]$
Now $\tan A-\tan B=\frac{\sin (A-B)}{\cos A\cos B}$
$\therefore p=\frac{\sin 18\theta }{\cos 27\theta \cos 9\theta }+\frac{\sin 6\theta }{\cos 9\theta \cos 3\theta }+\frac{\sin 2\theta }{\cos 3\theta \cos \theta }=\frac{2\sin 9\theta }{\cos 27\theta }+\frac{2\sin 3\theta }{\cos 9\theta }+\frac{2\sin \theta }{\cos 3\theta }$
or $p=2\left(q\right)$ or $\frac{p}{q}=2$