Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\tan \alpha .\cot \alpha = \frac{{\sin \alpha }}{{\cos \alpha }}.\frac{{\cos \alpha }}{{\sin \alpha }} = 1\\
\sin \alpha + \cos \alpha = m\\
\Leftrightarrow {\sin ^2}\alpha + 2\sin \alpha .\cos \alpha + {\cos ^2}\alpha = {m^2}\\
\Leftrightarrow 1 + 2\sin \alpha .\cos \alpha = {m^2}\\
\Leftrightarrow \sin \alpha .\cos \alpha = \frac{{{m^2} - 1}}{2}\\
{\tan ^2}\alpha + {\cot ^2}\alpha = \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} + \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }}\\
= \frac{{{{\sin }^4}\alpha + {{\cos }^4}\alpha }}{{{{\sin }^2}\alpha .{{\cos }^2}\alpha }} = \frac{{\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) - 2{{\sin }^2}\alpha .{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha .{{\cos }^2}\alpha }}\\
= \frac{{1 - 2{{\sin }^2}\alpha .{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha .{{\cos }^2}\alpha }} = \frac{1}{{{{\sin }^2}\alpha .{{\cos }^2}\alpha }} - 2\\
= \frac{1}{{{{\left( {\frac{{{m^2} - 1}}{2}} \right)}^2}}} - 2 = \frac{4}{{{{\left( {{m^2} - 1} \right)}^2}}} - 2
\end{array}\)
