Câu 6.29 trang 200 Sách BT 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) ${\tan ^3}\alpha  + {\cot ^3}\alpha .$

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)

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