Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\tan a = \dfrac{3}{5} \Leftrightarrow \dfrac{{\sin a}}{{\cos a}} = \dfrac{3}{5} \Leftrightarrow \sin a = \dfrac{3}{5}\cos a\\
a,\\
M = \dfrac{{\sin a.\cos a}}{{{{\sin }^2}a - {{\cos }^2}a}} = \dfrac{{\dfrac{3}{5}\cos a.\cos a}}{{{{\left( {\dfrac{3}{5}\cos a} \right)}^2} - {{\cos }^2}a}} = \dfrac{{\dfrac{3}{5}{{\cos }^2}a}}{{\dfrac{9}{{25}}{{\cos }^2}a - {{\cos }^2}a}} = \dfrac{{\dfrac{3}{5}{{\cos }^2}a}}{{ - \dfrac{{16}}{{25}}{{\cos }^2}a}} = - \dfrac{{15}}{{16}}\\
b,\\
P = \dfrac{{{{\sin }^3}a + {{\cos }^3}a}}{{2\sin a.{{\cos }^2}a + \cos a.{{\sin }^2}a}} = \dfrac{{{{\left( {\dfrac{3}{5}\cos a} \right)}^3} + {{\cos }^3}a}}{{2.\left( {\dfrac{3}{5}\cos a} \right).{{\cos }^2}a + \cos a.{{\left( {\dfrac{3}{5}\cos a} \right)}^2}}} = \dfrac{{\dfrac{{27}}{{125}}{{\cos }^3}a + {{\cos }^3}a}}{{\dfrac{6}{5}{{\cos }^3}a + \dfrac{9}{{25}}{{\cos }^3}a}} = \dfrac{{\dfrac{{152}}{{125}}{{\cos }^3}a}}{{\dfrac{{39}}{{25}}{{\cos }^2}a}} = \dfrac{{152}}{{195}}
\end{array}\)
