Cho: Tan$\alpha$ + cot$\alpha$ = 3 Tính C = $tan^4\alpha$ – $cot^4\alpha$ Biết: A = $tan^2\alpha$ + $cot^2\alpha$ = 11 B= tan$\alpha$ + cot$\alpha$ =

0 bình luận về “Cho: Tan$\alpha$ + cot$\alpha$ = 3 Tính C = $tan^4\alpha$ – $cot^4\alpha$ Biết: A = $tan^2\alpha$ + $cot^2\alpha$ = 11 B= tan$\alpha$ + cot$\alpha$ =”

1. Đáp án:

   $\begin{array}{l}
   Do:\cot a = \dfrac{1}{{\tan a}}\\
   Khi:\tan a + \cot a = 3\\
    \Leftrightarrow \tan a + \dfrac{1}{{\tan a}} = 3\\
    \Leftrightarrow {\tan ^2}a – 3\tan a + 1 = 0\\
    \Leftrightarrow \left[ \begin{array}{l}
   \tan a = \dfrac{{3 – \sqrt 5 }}{2};\cot a = \dfrac{{3 + \sqrt 5 }}{2}\\
   \tan a = \dfrac{{3 + \sqrt 5 }}{2};\cot a = \dfrac{{3 – \sqrt 5 }}{2}
   \end{array} \right.\\
    \Leftrightarrow C = {\tan ^4}a – {\cot ^4}a\\
    = {\left( {\dfrac{{3 \pm \sqrt 5 }}{2}} \right)^4} – {\left( {\dfrac{{3 \mp \sqrt 5 }}{2}} \right)^4}\\
    =  \pm 46,95
   \end{array}$

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