(1/(cos^2 10^o )) +(1/(sin^2 20^o )) + (1/(sin^2 40^o )) =?


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Question Number 95703 by i jagooll last updated on 27/May/20

$\frac{1}{\cos^{2} 10^{o}} + \frac{1}{\sin^{2} 20^{o}} + \frac{1}{\sin^{2} 40^{o}} = ?$
Commented by john santu last updated on 27/May/20

$this is equivalent to$ $\csc^{2} (\frac{π}{9}) + \csc^{2} (\frac{2 π}{9}) + \csc^{2} (\frac{4 π}{9})$
	Answered by john santu last updated on 27/May/20

	$\cos^{2} 10^{o} = \sin^{2} 80^{o}$ $\sin 60^{o} = 3 \sin 20^{o} - 4 \sin^{3} 20^{o} = \frac{\sqrt{3}}{2}$ $\sin 120^{o} = 3 \sin 40^{o} - 4 \sin^{3} 40^{o} = \frac{\sqrt{3}}{2}$ $\sin (- 240^{o}) = 3 \sin (- 80^{o}) - 4 \sin^{3} (- 80^{o}) = \frac{\sqrt{3}}{2}$ $so \sin 20^{o}; \sin 40^{o} & \sin (- 80^{o})$ $are the roots of cubic equation$ $of 3 x - {4 x}^{3} = \frac{\sqrt{3}}{2} or {4 x}^{3} - 3 x = \frac{\sqrt{3}}{2}$ $\Rightarrow x ({4 x}^{2} - 3) = \frac{\sqrt{3}}{2}; squaring both sides$ $x^{2} {({4 x}^{2} - 3)}^{2} = \frac{3}{4}, let x^{2} = t$ $\Rightarrow t {(4 t - 3)}^{2} = \frac{3}{4} \leftarrow are roots$ $\sin^{2} 20^{o}, \sin^{2} 40^{o}, \sin^{2} 80^{o}$ $t ({16 t}^{2} - 24 t + 9) - \frac{3}{4} = 0$ ${16 t}^{3} - {24 t}^{2} + 9 t - \frac{3}{4} = 0$ ${64 t}^{3} - {96 t}^{2} + 36 t - 3 = 0$ $put t = \frac{1}{y} \Rightarrow - {3 t}^{3} + {36 t}^{2} - 96 t + 64 = 0$ $are roots \frac{1}{\sin^{2} 20^{o}}, \frac{1}{\sin^{2} 40^{o}} and \frac{1}{\sin^{2} 80^{o}}$ $hence by {vieta}^{'} s rule we get$ $\frac{1}{\sin^{2} 20^{o}} + \frac{1}{\sin^{2} 40^{o}} + \frac{1}{\sin^{2} 80^{o}} = \frac{- 36}{- 3} = 12$
	Commented by i jagooll last updated on 27/May/20

	$greatt$
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