cosec⁡360∘7+cosec⁡540∘7=

let α=180∘7⇒ 180∘=7α⇒ 3α=180∘−4α⇒ sin⁡3α=sin⁡4α----i∴ cosec⁡360∘7+cosec⁡540∘7 =cosec⁡2α+cosec⁡3α =sin⁡3α+sin⁡2αsin⁡3αsin⁡2α =sin⁡4α+sin⁡2αsin⁡3α⋅sin⁡2α =2sin⁡3αcos⁡αsin⁡3α⋅sin⁡2α from (i) =2cos⁡α2sin⁡α⋅cos⁡α =cosec⁡α =cosec⁡180∘7

cosec⁡360∘7+cosec⁡540∘7=

let α=180∘7⇒ 180∘=7α⇒ 3α=180∘−4α⇒ sin⁡3α=sin⁡4α----i∴ cosec⁡360∘7+cosec⁡540∘7 =cosec⁡2α+cosec⁡3α =sin⁡3α+sin⁡2αsin⁡3αsin⁡2α =sin⁡4α+sin⁡2αsin⁡3α⋅sin⁡2α =2sin⁡3αcos⁡αsin⁡3α⋅sin⁡2α from (i) =2cos⁡α2sin⁡α⋅cos⁡α =cosec⁡α =cosec⁡180∘7

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$cosec \frac{360^{\circ}}{7} + cosec \frac{540^{\circ}}{7} =$

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$cosec \frac{180^{\circ}}{7}$

$cosec \frac{90^{\circ}}{7}$

$\sec \frac{180^{\circ}}{7}$

$\sec \frac{90^{\circ}}{7}$

answer is A.

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Detailed Solution

$let α = \frac{180^{\circ}}{7}$

$\begin{matrix} \Rightarrow 180^{\circ} = 7 α \\ \Rightarrow 3 α = 180^{\circ} - 4 α \\ \Rightarrow \sin 3 α = \sin 4 α - - - - (i) \\ ∴ cosec \frac{360^{\circ}}{7} + cosec \frac{540^{\circ}}{7} \\ = cosec 2 α + cosec 3 α \\ = \frac{\sin 3 α + \sin 2 α}{\sin 3 αsin 2 α} \\ = \frac{\sin 4 α + \sin 2 α}{\sin 3 α \cdot \sin 2 α} \\ = \frac{2 \sin 3 αcos α}{\sin 3 α \cdot \sin 2 α} from (i) \\ = \frac{2 \cos α}{2 \sin α \cdot \cos α} \\ = cosec α \end{matrix}$