જ્યારે frac{sin 9 theta}{cos 27 theta}+frac{s | vedclass

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(C) આપેલ પદાવલિ: $\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 	h

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$\frac{1}{2}$

Vedclass · Answer

(C) આપેલ પદાવલિ: $\frac{\sin 9 	heta}{\cos 27 	heta}+\frac{\sin 3 	heta}{\cos 9 	heta}+\frac{\sin 	heta}{\cos 3 	heta}=k(	an 27 	heta-	an 	heta)$.$	heta = \frac{\pi}{3}$ મૂકતા:$\frac{\sin 9(\frac{\pi}{3})}{\cos 27(\frac{\pi}{3})} + \frac{\sin 3(\frac{\pi}{3})}{\cos 9(\frac{\pi}{3})} + \frac{\sin(\frac{\pi}{3})}{\cos 3(\frac{\pi}{3})} = k(	an 27(\frac{\pi}{3}) - 	an(\frac{\pi}{3}))$.$\sin(3\pi) = 0$,$\sin(\pi) = 0$,અને $\cos(9\pi) = -1$,$\cos(3\pi) = -1$,$\cos(\pi) = -1$ હોવાથી:$0 + 0 + \frac{\sqrt{3}/2}{-1} = k(0 - \sqrt{3})$.$-\frac{\sqrt{3}}{2} = -k\sqrt{3}$.તેથી,$k = \frac{1}{2}$.

જ્યારે $\frac{\sin 9 \theta}{\cos 27 \theta}+\frac{\sin 3 \theta}{\cos 9 \theta}+\frac{\sin \theta}{\cos 3 \theta}=k(\tan 27 \theta-\tan \theta)$ વ્યાખ્યાયિત હોય,ત્યારે $k=$

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