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