Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\dfrac{{\tan x - \sin x}}{{{{\sin }^3}x}} = \dfrac{{\dfrac{{\sin x}}{{\cos x}} - \sin x}}{{{{\sin }^3}x}} = \dfrac{{\dfrac{1}{{\cos x}} - 1}}{{{{\sin }^2}x}} = \dfrac{{\dfrac{{1 - \cos x}}{{\cos x}}}}{{1 - {{\cos }^2}x}} = \dfrac{{1 - \cos x}}{{\cos x\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)}} = \dfrac{1}{{\cos x\left( {1 + \cos x} \right)}}\\
b,\\
\cot a = \dfrac{1}{{\tan a}} = - \dfrac{1}{{\sqrt 3 }}\\
\tan a = - \sqrt 3 \Leftrightarrow \dfrac{{\sin a}}{{\cos a}} = - \sqrt 3 \Leftrightarrow \sin a = - \sqrt 3 \cos a\\
{\sin ^2}a + {\cos ^2}a = 1\\
\Leftrightarrow {\left( { - \sqrt 3 \cos a} \right)^2} + {\cos ^2}a = 1\\
\Leftrightarrow {\cos ^2}a = \dfrac{1}{4} \Rightarrow {\sin ^2}a = \dfrac{3}{4}\\
P = \dfrac{{2\cot a - 3{{\cos }^2}a}}{{{{\sin }^2}a - 2}} = \dfrac{{2.\dfrac{{ - 1}}{{\sqrt 3 }} - 3.\dfrac{1}{4}}}{{\dfrac{3}{4} - 2}} = \dfrac{{9 + 8\sqrt 3 }}{{15}}\\
c,\\
P = \left( {\dfrac{{1 + \tan a}}{{1 + \sin 2a}}} \right).\cos a = \left( {\dfrac{{1 + \dfrac{{\sin a}}{{\cos a}}}}{{1 + 2\sin a.\cos a}}} \right).\cos a\\
= \dfrac{{\cos a + \sin a}}{{1 + 2\sin a.\cos a}} = \dfrac{{\sin a + \cos a}}{{\left( {{{\sin }^2}a + {{\cos }^2}a} \right) + 2\sin a.\cos a}}\\
= \dfrac{{\sin a + \cos a}}{{{{\left( {\sin a + \cos a} \right)}^2}}} = \dfrac{1}{{\sin a + \cos a}} = - 1 - \sqrt 3
\end{array}\)
