Question-132205

Question Number 132205 by Arijit last updated on 12/Feb/21

Commented by Arijit last updated on 12/Feb/21

Please Help me to solve this \dots . .

Answered by mathmax by abdo last updated on 13/Feb/21

A = \tan^{2} (\frac{π}{16}) + \tan^{2} (\frac{7 π}{16}) + \tan^{2} (\frac{2 π}{16}) + \tan^{2} (\frac{6 π}{16}) + \tan^{2} (\frac{3 π}{16}) + \tan^{2} (\frac{5 π}{16})

+ \tan^{2} (\frac{4 π}{16}) = \tan^{2} (\frac{π}{16}) + \tan^{2} (\frac{π}{2} - \frac{π}{16}) + \tan^{2} (\frac{2 π}{16}) + \tan^{2} (\frac{π}{2} - \frac{2 π}{16})

+ \tan^{2} (\frac{3 π}{16}) + \tan^{2} (\frac{π}{2} - \frac{3 π}{16}) + \tan^{2} (\frac{π}{4})

= \tan^{2} (\frac{π}{16}) + \frac{1}{\tan^{2} (\frac{π}{16})} + \tan^{2} (\frac{2 π}{16}) + \frac{1}{\tan^{2} (\frac{2 π}{16})} + \tan^{2} (\frac{3 π}{16}) + \frac{1}{\tan^{2} (\frac{3 π}{16})}

= \tan^{2} (\frac{π}{16}) + \frac{1}{\tan^{2} (\frac{π}{16})} + {(\sqrt{2} - 1)}^{2} + \frac{1}{{(\sqrt{2} - 1)}^{2}} + \tan^{2} (\frac{3 π}{16}) + \frac{1}{\tan^{2} (\frac{3 π}{16})}

let \tan (\frac{π}{16}) = x we have tanx = \frac{2 \tan (\frac{x}{2})}{1 - \tan^{2} (\frac{x}{2})} \Rightarrow for x = \frac{π}{8} we get

\sqrt{2} - 1 = \frac{2 x}{1 - x^{2}} \Rightarrow (\sqrt{2} - 1) (1 - x^{2}) - 2 x = 0 \Rightarrow

(\sqrt{2} - 1) + (1 - \sqrt{2}) x^{2} - 2 x = 0 \Rightarrow (1 - \sqrt{2}) x^{2} - 2 x + \sqrt{2} - 1 = 0

Δ^{^{'}} = 1 - (1 - \sqrt{2}) (\sqrt{2} - 1) = 1 + {(\sqrt{2} - 1)}^{2} = 1 + 3 - 2 \sqrt{2} = 4 - 2 \sqrt{2}

x_{1} = \frac{1 + \sqrt{4 - 2 \sqrt{2}}}{1 - \sqrt{2}} < 0 and x_{2} = \frac{1 - \sqrt{4 - 2 \sqrt{2}}}{1 - \sqrt{2}} = \frac{- 1 + \sqrt{4 - 2 \sqrt{2}}}{\sqrt{2} - 1} > 0 \Rightarrow

\tan (\frac{π}{16}) = \frac{\sqrt{4 - 2 \sqrt{2}} - 1}{\sqrt{2} - 1} = (\sqrt{2} + 1) (\sqrt{4 - 2 \sqrt{2}} - 1) \Rightarrow

\tan^{2} (\frac{π}{16}) = (3 + 2 \sqrt{2}) (4 - 2 \sqrt{2} + 1 - 2 \sqrt{4 - 2 \sqrt{2}})

= (3 + 2 \sqrt{2}) (5 - 2 \sqrt{2} - 2 \sqrt{4 - 2 \sqrt{2}}) rest to find \tan (\frac{3 π}{16}) \dots be continued \dots

Commented by Arijit last updated on 14/Feb/21

T h a n k y o u v e r y m u c h

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