$\alpha$ là một góc cố định cho trước. Tìm giá trị nhỏ nhất của hàm số :                              $y = tan ^2(x + \alpha ) +tan^2(x – \alpha )$

Ta có
$y = tan^2(x + \alpha ) + tan^2(x – \alpha ) = \frac{{{{\sin }^2}(x + \alpha )}}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}(x + \alpha )}} + \frac{{{{\sin }^2}(x – \alpha )}}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}(x – \alpha )}}$
    $ = \frac{{{{\sin }^2}(x + \alpha )c{\rm{o}}{{\rm{s}}^{\rm{2}}}(x – \alpha ) + {{\sin }^2}(x – \alpha )c{\rm{o}}{{\rm{s}}^{\rm{2}}}(x + \alpha )}}{{c{\rm{o}}{{\rm{s}}^{\rm{2}}}(x + \alpha )c{\rm{o}}{{\rm{s}}^{\rm{2}}}(x – \alpha )}}$
    $ = \frac{{{{(\sin 2x + \sin 2\alpha )}^2}/4 + {{(\sin 2x – \sin 2\alpha )}^2}/4}}{{{{(co{\mathop{\rm s}\nolimits} 2x + co{\mathop{\rm s}\nolimits} 2\alpha )}^2}/4}}$
    $ = \frac{{2({{\sin }^2}2x + {{\sin }^2}2\alpha )}}{{{{(co{\mathop{\rm s}\nolimits} 2x + co{\mathop{\rm s}\nolimits} 2\alpha )}^2}}} = \frac{{2(1 – co{{\mathop{\rm s}\nolimits} ^2}2x + {{\sin }^2}2\alpha )}}{{{{(co{\mathop{\rm s}\nolimits} 2x + co{\mathop{\rm s}\nolimits} 2\alpha )}^2}}}$
$\Rightarrow \min y = \left\{ \begin{array}{l}
{\sin ^2}2\alpha /{(1 + co{\mathop{\rm s}\nolimits} 2\alpha )^2}nếu  cos2\alpha  \ge {\rm{0 (khi  cos2x  = 1)}}\\
{\sin ^2}2\alpha /{(1 – co{\mathop{\rm s}\nolimits} 2\alpha )^2}nếu  cos2\alpha  \le {\rm{0 (khi  cos2x  = – 1)}}
\end{array} \right.$
               $ = \left\{ \begin{array}{l}
2tan^2\alpha ,nếu  cos2\alpha  \ge {\rm{0 (khi  cos2x  = 1)}}\\
2\cot ^2\alpha ,nếu  cos2\alpha  \le {\rm{0 (khi  cos2x  = – 1)}}
\end{array} \right.$