When frac{sin 9 theta}{cos 27 theta}+frac{sin | vedclass

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Question

(C) Given the expression: $\frac{\sin 9 	heta}{\cos 27 	heta}+\frac{\sin 3 	heta}{\cos 9 	heta}+\frac{\sin 	heta}{\cos 3 	heta}=k(	an 27 	heta

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) Given the expression: $\frac{\sin 9 	heta}{\cos 27 	heta}+\frac{\sin 3 	heta}{\cos 9 	heta}+\frac{\sin 	heta}{\cos 3 	heta}=k(	an 27 	heta-	an 	heta)$.Substitute $	heta = \frac{\pi}{3}$ in the expression:$\frac{\sin 9(\frac{\pi}{3})}{\cos 27(\frac{\pi}{3})} + \frac{\sin 3(\frac{\pi}{3})}{\cos 9(\frac{\pi}{3})} + \frac{\sin(\frac{\pi}{3})}{\cos 3(\frac{\pi}{3})} = k(	an 27(\frac{\pi}{3}) - 	an(\frac{\pi}{3}))$.Since $\sin(3\pi) = 0$,$\sin(\pi) = 0$,and $\cos(9\pi) = -1$,$\cos(3\pi) = -1$,$\cos(\pi) = -1$:$0 + 0 + \frac{\sqrt{3}/2}{-1} = k(0 - \sqrt{3})$.$-\frac{\sqrt{3}}{2} = -k\sqrt{3}$.Therefore,$k = \frac{1}{2}$.

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 $k=$

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