cot^2 ((π/7))+cot^2 (((2π)/7))+cot^2 (((3π)/7))=?


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Question Number 128582 by john_santu last updated on 08/Jan/21

$\cot^{2} (\frac{π}{7}) + \cot^{2} (\frac{2 π}{7}) + \cot^{2} (\frac{3 π}{7}) = ?$
	Answered by liberty last updated on 08/Jan/21

	$T = \cot^{2} (\frac{π}{7}) + \cot^{2} (\frac{2 π}{7}) + \cot^{2} (\frac{3 π}{7})$ $T = \csc^{2} (\frac{π}{7}) + \csc^{2} (\frac{2 π}{7}) + \csc^{2} (\frac{3 π}{7}) - 3$ $T = \frac{1}{\sin^{2} (\frac{π}{7})} + \frac{1}{\sin^{2} (\frac{2 π}{7})} + \frac{1}{\sin^{2} (\frac{3 π}{7})} - 3$ $T = \frac{2}{1 - \cos (\frac{2 π}{7})} + \frac{2}{1 - \cos (\frac{4 π}{7})} + \frac{2}{1 - \cos (\frac{6 π}{7})} - 3$ $T = \frac{2}{1 + \cos (\frac{π}{7})} + \frac{2}{1 + \cos (\frac{3 π}{7})} + \frac{2}{1 + \cos (\frac{5 π}{7})} - 3$ $T = \frac{2 [(1 + \cos \frac{π}{7}) (1 + \cos \frac{3 π}{7}) + (1 + \cos \frac{π}{7}) (1 + \cos \frac{5 π}{7}) + (1 + \cos \frac{3 π}{7}) (1 + \cos \frac{5 π}{7})]}{(1 + \cos \frac{π}{7}) (1 + \cos \frac{3 π}{7}) (1 + \cos \frac{5 π}{7})} - 3$ $T = \frac{2 [1 + \cos \frac{π}{7} + \cos \frac{3 π}{7} + \cos \frac{π}{7} . \cos \frac{3 π}{7} + 1 + \cos \frac{π}{7} + \cos \frac{5 π}{7} + \cos \frac{π}{7} . \cos \frac{5 π}{7} + 1 + \cos \frac{3 π}{7} + \cos \frac{5 π}{7} + \cos \frac{3 π}{7} . \cos \frac{5 π}{7}]}{(1 + \cos \frac{5 π}{7}) (1 + \cos \frac{π}{7} + \cos \frac{3 π}{7} + \cos \frac{π}{7} . \cos \frac{3 π}{7})} - 3$ $T = 8 - 3 = 5$
	Commented by Study last updated on 10/Jan/21

	$w h y 1 - c o s (\frac{2 π}{7}) = 1 + c o s (\frac{π}{7})$ $1 - c o s (\frac{4 π}{7}) = 1 + c o s (\frac{3 π}{7})$ $1 - c o s \frac{6 π}{7} = 1 + c o s \frac{5 π}{7} ? ? ?$
	Commented by liberty last updated on 13/Jan/21

	$1 - \cos (\frac{6 π}{7}) = 1 + \cos (\frac{π}{7})$ $1 - \cos (\frac{4 π}{7}) = 1 + \cos (\frac{3 π}{7})$ $1 - \cos (\frac{2 π}{7}) = 1 + \cos (\frac{5 π}{7})$
	Commented by Adel last updated on 14/Jan/21

	Please help improve this article or section by expanding it
	Commented by Adel last updated on 14/Jan/21

	$1 - \cos (\frac{2 π}{7}) = 1 - \cos (π - \frac{5 π}{7}) = 1 - (- \cos \frac{5 π}{7}) = 1 + \cos (\frac{5 π}{7})$ $1 - \cos (\frac{4 π}{7}) = 1 - \cos (π - \frac{5 π}{7}) = 1 - {- \cos (\frac{5 π}{7})} = 1 + \cos (\frac{5 π}{7})$ $1 - \cos (\frac{6 π}{7}) = 1 - \cos (π - \frac{π}{7}) = 1 - {- \cos (\frac{π}{7})} = 1 + \cos (\frac{π}{7})$ $Solve Adil Rahman Danish$
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