Câu 6.29 trang 200 SBT Đại số 10 Nâng cao

Cho \(\tan \alpha  + \cot \alpha  = m\), hãy tính theo \(m\)

a) \({\tan ^2}\alpha  + {\cot ^2}\alpha ;\)

b) \(\left| {\tan \alpha  - \cot \alpha } \right|;\)

c) \({\tan ^3}\alpha  + {\cot ^3}\alpha .\)

Giải:

Cho \(\tan \alpha  + \cot \alpha  = m\), ta có:

a)

\(\begin{array}{l}{\tan ^2}\alpha  + {\cot ^2}\alpha \\ = {\left( {{{\tan }^2}\alpha  + \cot \alpha } \right)^2} - 2\tan \alpha \cot \alpha \\ = {m^2} - 2\end{array}\)

b)

\(\begin{array}{l}{\left( {\tan \alpha  - \cot \alpha } \right)^2}\\ = {\tan ^2}\alpha  + {\cot ^2}\alpha  - 2\tan \alpha \cot \alpha \\ = {m^2} - 4\end{array}\)

Vậy \(\left| {\tan \alpha  - \cot \alpha } \right| = \sqrt {{m^2} - 4} \) (để ý rằng, do \(\tan \alpha .\cot \alpha  = 1\) nên \(\left| {\tan \alpha  + \cot \alpha } \right| \ge 2\), từ đó \({m^2} \ge 4\))

c)

\(\begin{array}{l}{\tan ^3}\alpha  + {\cot ^3}\alpha \\ = {\left( {\tan \alpha  + \cot \alpha } \right)^3} - 3\tan \alpha \cot \alpha \left( {\tan \alpha  + \cot \alpha } \right)\\ = {m^3} - 3m\end{array}\)

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