operatorname{cosec} 48^{circ}+operatorname{co | vedclass

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Question

(C) Let $S = \operatorname{cosec} 48^{\circ} + \operatorname{cosec} 96^{\circ} + \operatorname{cosec} 192^{\circ} + \operatorname{cosec} 384^{\circ}$.

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Vedclass · Answer

(C) Let $S = \operatorname{cosec} 48^{\circ} + \operatorname{cosec} 96^{\circ} + \operatorname{cosec} 192^{\circ} + \operatorname{cosec} 384^{\circ}$.Using the identity $\operatorname{cosec} 	heta = \frac{1}{\sin 	heta}$,we have:$S = \frac{1}{\sin 48^{\circ}} + \frac{1}{\sin 96^{\circ}} + \frac{1}{\sin 192^{\circ}} + \frac{1}{\sin 384^{\circ}}$.Note that $\sin 192^{\circ} = \sin(180^{\circ} + 12^{\circ}) = -\sin 12^{\circ}$ and $\sin 384^{\circ} = \sin(360^{\circ} + 24^{\circ}) = \sin 24^{\circ}$.Also,$\sin 96^{\circ} = \sin(180^{\circ} - 84^{\circ}) = \sin 84^{\circ}$.Using the identity $\operatorname{cosec} 	heta + \operatorname{cosec}(180^{\circ} + 	heta) = \operatorname{cosec} 	heta - \operatorname{cosec} 	heta = 0$ is not directly applicable here.However,using the identity $\operatorname{cosec} 	heta + \operatorname{cosec}(2	heta) = \frac{1}{\sin 	heta} + \frac{1}{2 \sin 	heta \cos 	heta} = \frac{2 \cos 	heta + 1}{2 \sin 	heta \cos 	heta} = \frac{\cos 	heta + \cos 	heta + 1}{\sin 2	heta}$.Applying the property $\operatorname{cosec} 	heta + \operatorname{cosec}(180^{\circ} + 	heta) = 0$ is incorrect. Let us use the identity $\operatorname{cosec} 	heta = \cot(	heta/2) - \cot 	heta$.Then $S = (\cot 24^{\circ} - \cot 48^{\circ}) + (\cot 48^{\circ} - \cot 96^{\circ}) + (\cot 96^{\circ} - \cot 192^{\circ}) + (\cot 192^{\circ} - \cot 384^{\circ})$.This is a telescoping sum: $S = \cot 24^{\circ} - \cot 384^{\circ}$.Since $\cot 384^{\circ} = \cot(360^{\circ} + 24^{\circ}) = \cot 24^{\circ}$,we get $S = \cot 24^{\circ} - \cot 24^{\circ} = 0$.

$\operatorname{cosec} 48^{\circ}+\operatorname{cosec} 96^{\circ}+\operatorname{cosec} 192^{\circ}+\operatorname{cosec} 384^{\circ}=$

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