prove-that-tan-2-36-tan-2-72-5-

Question Number 143320 by mathdanisur last updated on 12/Jun/21

p r o v e t h a t : {t a n}^{2} 36 ° + {t a n}^{2} 72 ° = 5

Answered by Ar Brandon last updated on 12/Jun/21

\tan^{2} 36 ° + \tan^{2} 72 ° = \tan^{2} 36 ° + \cot^{2} 18 °

= {(\frac{\sqrt{10 - 2 \sqrt{5}}}{\sqrt{5} + 1})}^{2} + {(\frac{\sqrt{10 + 2 \sqrt{5}}}{\sqrt{5} - 1})}^{2}

= \frac{10 - 2 \sqrt{5}}{6 + 2 \sqrt{5}} + \frac{10 + 2 \sqrt{5}}{6 - 2 \sqrt{5}}

= \frac{(80 - 32 \sqrt{5}) + (80 + 32 \sqrt{5})}{16}

= \frac{160}{16} = 10

Commented by mathdanisur last updated on 13/Jun/21

t h a n k s S i r . .

Answered by MJS_new last updated on 13/Jun/21

{it}^{'} s wrong

Answered by MJS_new last updated on 13/Jun/21

\tan^{2} x + \tan^{2} 2 x = 10

\tan x = t

\frac{t^{2} (t^{4} - 2 t^{2} + 5)}{{(t^{2} + 1)}^{2}} = 10

t^{6} - 12 t^{4} + 25 t^{2} - 10 = 0

(t^{2} - 2) (t^{4} - 10 t^{2} + 5) = 0

t = \pm \sqrt{2} \lor t = \pm \sqrt{5 - 2 \sqrt{5}} \lor t = \pm \sqrt{5 + 2 \sqrt{5}}

t = \tan x \Leftrightarrow x = n π + \arctan t

for 0 ⩽ x < 90 ° we get

x = 36 ° \lor x \approx 54.74 ° \lor x = 72 °

Commented by mathdanisur last updated on 13/Jun/21

S i r t h a t i s i t c a n n o t b e 5 . ?

Commented by MJS_new last updated on 13/Jun/21

no . just type it in any calculator .

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