Question

If $k_1=\tan 27\theta -\tan \theta$ and $k_2=\frac{\sin \theta }{\cos 3\theta }+\frac{\sin 3\theta }{\cos 9\theta }+\frac{\sin 9\theta }{\cos 27\theta }$ then, find the relation between $k_1$ and $k_2$.

Answer 1

We have $k_1=\tan 27\theta -\tan \theta$

$=(\tan 27\theta -\tan 9\theta )+(\tan 9\theta -\tan 3\theta )+(\tan 3\theta -\tan \theta )$

Consider $(\tan 27\theta -\tan 9\theta )=\frac{\sin 3\theta }{\cos 3\theta }-\frac{\sin \theta }{\cos \theta }$

$=\frac{\sin 3\theta \cos \theta -\cos 3\theta \sin \theta }{\cos 3\theta \cos \theta }$

$=\frac{\sin (3\theta -\theta )}{\cos 3\theta \cos \theta }$ .................. using $\sin A\cos B-\cos A\sin B=\sin (A-B)$

$=\frac{\sin 2\theta }{\cos 3\theta \cos \theta }$

$=\frac{2\sin \theta \cos \theta }{\cos 3\theta \cos \theta }$

$=\frac{2\sin \theta }{\cos 3\theta }$

Consider $(\tan 9\theta -\tan 3\theta )=\frac{\sin 9\theta }{\cos 9\theta }-\frac{\sin 3\theta }{\cos 3\theta }$

$=\frac{\sin 9\theta \cos 3\theta -\cos 9\theta \sin 3\theta }{\cos 9\theta \cos 3\theta }$

$=\frac{\sin (9\theta -3\theta )}{\cos 9\theta \cos 3\theta }$ ................. using $\sin A\cos B-\cos A\sin B=\sin (A-B)$

$=\frac{\sin 6\theta }{\cos 9\theta \cos 3\theta }$

$=\frac{2\sin 3\theta \cos 3\theta }{\cos 9\theta \cos \theta }$

$=\frac{2\sin 3\theta }{\cos 9\theta }$

Consider $=(\tan 27\theta -\tan 9\theta )=\frac{\sin 27\theta }{\cos 27\theta }-\frac{\sin 9\theta }{\cos 9\theta }$

$=\frac{\sin 27\theta \cos 9\theta -\cos 27\theta \sin 9\theta }{\cos 27\theta \cos 9\theta }$

$=\frac{\sin (27\theta -9\theta )}{\cos 27\theta \cos 9\theta }$ .................... using $\sin A\cos B-\cos A\sin B=\sin (A-B)$

$=\frac{\sin 18\theta }{\cos 27\theta \cos 9\theta }$

$=\frac{2\sin 9\theta \cos 9\theta }{\cos 27\theta \cos 9\theta }$

$=\frac{2\sin 9\theta }{\cos 27\theta }$

$\therefore k_1=(\tan 27\theta -\tan 9\theta )+(\tan 9\theta -\tan 3\theta )+(\tan 3\theta -\tan \theta )$

$=\frac{2\sin \theta }{\cos 3\theta }+\frac{2\sin 3\theta }{\cos 9\theta }+\frac{2\sin 9\theta }{\cos 27\theta }$

$=2(\frac{\sin \theta }{\cos 3\theta }+\frac{\sin 3\theta }{\cos 9\theta }+\frac{\sin 9\theta }{\cos 27\theta })=2k_2$