Question

If $k_1=tan27\theta -tan\theta$ and $k_2=\frac{sin\theta }{cos3\theta }+\frac{sin3\theta }{cos9\theta }+\frac{sin9\theta }{cos27\theta }$
then,

• A. $k_1=k_2$
• B. $k_1=2k_2$
• C. $k_1+k_2=2$
• D. $k_2=2k_1$

Answer 1

Answer: (B)

$k_1=2k_2$

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

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

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

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

Similarly,

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

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

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

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

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

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

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