If α=π7 then |tan⁡αtan⁡2α+tan⁡2αtan⁡4α+tan⁡4αtan⁡α|=

α=π7⇒α+2α+4α=πtan⁡αtan⁡2α+tan⁡2αtan⁡4α+tan⁡4αtan⁡αsin⁡αsin⁡2αcos⁡4α+sin⁡2αsin⁡4αcos⁡α+sin⁡4αsin⁡αcos⁡2αcos⁡αcos⁡2αcos⁡4α=cos⁡αcos⁡2αcos⁡4α−cos⁡(α+2α+4α)cos⁡αcos⁡2αcos⁡4α=1+sec⁡αsec⁡2αsec⁡4α Now, cos⁡αcos⁡2αcos⁡4α=sin⁡8α23sin⁡α =−sin⁡α8sin⁡α=−18 [∵7α=π]hence |tan⁡αtan⁡2α+tan⁡2αtan⁡4α+tan⁡4αtan⁡α|=|1−8|=7

If α=π7 then |tan⁡αtan⁡2α+tan⁡2αtan⁡4α+tan⁡4αtan⁡α|=

α=π7⇒α+2α+4α=πtan⁡αtan⁡2α+tan⁡2αtan⁡4α+tan⁡4αtan⁡αsin⁡αsin⁡2αcos⁡4α+sin⁡2αsin⁡4αcos⁡α+sin⁡4αsin⁡αcos⁡2αcos⁡αcos⁡2αcos⁡4α=cos⁡αcos⁡2αcos⁡4α−cos⁡(α+2α+4α)cos⁡αcos⁡2αcos⁡4α=1+sec⁡αsec⁡2αsec⁡4α Now, cos⁡αcos⁡2αcos⁡4α=sin⁡8α23sin⁡α =−sin⁡α8sin⁡α=−18 [∵7α=π]hence |tan⁡αtan⁡2α+tan⁡2αtan⁡4α+tan⁡4αtan⁡α|=|1−8|=7

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If $α = \frac{π}{7}$ then $| \tan α \tan 2 α + \tan 2 α \tan 4 α + \tan 4 α \tan α | =$

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Detailed Solution

$\begin{array}{l} α = \frac{π}{7} \Rightarrow α + 2 α + 4 α = π \\ \begin{array}{c} \tan α \tan 2 α + \tan 2 α \tan 4 α + \tan 4 α \tan α \\ \sin α \sin 2 α \cos 4 α + \sin 2 α \sin 4 α \cos α + \sin 4 α \sin α \cos 2 α \\ \cos α \cos 2 α \cos 4 α \end{array} \\ = \frac{\cos α \cos 2 α \cos 4 α - \cos (α + 2 α + 4 α)}{\cos α \cos 2 α \cos 4 α} \\ = 1 + \sec α \sec 2 α \sec 4 α \end{array}$
$\begin{aligned} Now, \cos α \cos 2 α \cos 4 α = \frac{\sin 8 α}{2^{3} \sin α} \\ = \frac{- \sin α}{8 \sin α} = - \frac{1}{8} [∵ 7 α = π] \end{aligned}$
hence
$\begin{array}{l} | \tan αtan 2 α + \tan 2 αtan 4 α + \tan 4 αtan α | \\ = | 1 - 8 | = 7 \end{array}$