Discussion Forum

1)

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

 $(\tan 27 \theta -\tan \theta)$  is defined , then k=

Answer:

Option C

Explanation:

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

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

=$\frac{\sin 18 \theta+\sin 30 \theta-\sin 24 \theta}{2\cos 27 \theta\cos 9 \theta}+\frac{\sin  \theta}{\cos 3 \theta}$

=$\frac{\sin 21 \theta+\sin 15 \theta+\sin 33 \theta+\sin 27 \theta-\sin 27 \theta-\sin 21 \theta +4\sin \theta\cos 9 \theta \cos 27 \theta}{4\cos 3 \theta\cos 9 \theta \cos 27 \theta}$

=$\frac{\sin 15 \theta+\sin 33 \theta +4\sin \theta\cos 9 \theta \cos 27 \theta}{4\cos 3 \theta\cos 9 \theta \cos 27 \theta}$

=$\frac{2\sin 24 \theta \cos 9 \theta+4\sin \theta\cos 9 \theta \cos 27 \theta}{4\cos 3 \theta\cos 9 \theta \cos 27 \theta}$

=$\frac{\sin 24 \theta+2\sin \theta\cos 27 \theta}{2\cos 3 \theta\cos 27 \theta}$

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

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

=$\frac{1}{2}\frac{\sin 27 \theta}{\cos 27 \theta}-\frac{1}{2}\left(\frac{\sin 3 \theta-2\sin \theta}{\cos 3 \theta}\right)$

=$\frac{1}{2}\tan 27 \theta-\frac{1}{2}\frac{3\sin  \theta-4\sin^{3} \theta-2\sin \theta}{\cos  \theta(4\cos^{2}\theta-3)}$

=$\frac{1}{2}\tan 27 \theta-\tan \theta$

So,k=$\frac{1}{2}$