cot2π7+cot22π7+cot23π7 equals

Let 7θ=nπ, where n=1,2,3 ∴tan⁡θ can take the values tan⁡π7,tan⁡2π7,tan⁡3π7 4θ=nπ−3θ ⇒tan⁡4θ=tan⁡(nπ−3θ)=−tan⁡3θ⇒2tan⁡2θ1−tan2⁡2θ=−3tan⁡θ−tan3⁡θ1−3tan2⁡θ⇒4tan⁡θ1−tan2⁡θ1−tan2⁡θ2−4tan2⁡θ=−3tan⁡θ−tan3⁡θ1−3tan2⁡θ⇒tan2⁡θ3−21tan2⁡θ2+35tan2⁡θ−7=0 ⇒7cot2⁡θ3−35cot2⁡θ2+21cot2⁡θ−1=0 Sum of the roots=cot2⁡π7+cot2⁡2π7+cot2⁡3π7=357=5

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$\cot^{2} \frac{π}{7} + c o t^{2} \frac{2 π}{7} + \cot^{2} \frac{3 π}{7}$ equals

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answer is 5.

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Detailed Solution

$Let 7 θ = n π, where n = 1, 2, 3 ∴ \tan θ can take the values \tan \frac{π}{7}, \tan \frac{2 π}{7}, \tan \frac{3 π}{7} 4 θ = n π - 3 θ \begin{array}{l} \Rightarrow \tan 4 θ = \tan (n π - 3 θ) = - \tan 3 θ \\ \Rightarrow \frac{2 \tan 2 θ}{1 - \tan^{2} 2 θ} = - \frac{3 \tan θ - \tan^{3} θ}{1 - 3 \tan^{2} θ} \\ \Rightarrow \frac{4 \tan θ (1 - \tan^{2} θ)}{{(1 - \tan^{2} θ)}^{2} - 4 \tan^{2} θ} = - \frac{3 \tan θ - \tan^{3} θ}{1 - 3 \tan^{2} θ} \\ \Rightarrow {(\tan^{2} θ)}^{3} - 21 {(\tan^{2} θ)}^{2} + 35 \tan^{2} θ - 7 = 0 \Rightarrow 7 {(\cot^{2} θ)}^{3} - 35 {(\cot^{2} θ)}^{2} + 21 \cot^{2} θ - 1 = 0 Sum of the roots= \cot^{2} \frac{π}{7} + \cot^{2} \frac{2 π}{7} + \cot^{2} \frac{3 π}{7} = \frac{35}{7} = 5 \end{array}$