Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2,\\
90^\circ < x < 180^\circ \Rightarrow \left\{ \begin{array}{l}
\sin x > 0\\
\cos x < 0
\end{array} \right.\\
{\sin ^2}x + {\cos ^2}x = 1\\
\Leftrightarrow {\left( {\dfrac{1}{3}} \right)^2} + {\cos ^2}x = 1\\
\Leftrightarrow {\cos ^2}x = \dfrac{8}{9}\\
\cos x < 0 \Rightarrow \cos x = - \dfrac{{2\sqrt 2 }}{3}\\
\Rightarrow \tan x = \dfrac{{\sin x}}{{\cos x}} = - \dfrac{1}{{2\sqrt 2 }}\\
\cot x = \dfrac{{\cos x}}{{\sin x}} = - 2\sqrt 2 \\
M = \dfrac{{\sqrt 2 \cos x + {{\sin }^2}x}}{{\sqrt 2 \tan x + {{\cot }^2}x}} = \dfrac{{ - 22}}{{135}}\\
b,\\
\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = 2R\\
\Rightarrow a.\sin B = b.\sin A\\
\Leftrightarrow \dfrac{{\sin A}}{{\sin B}} = \dfrac{a}{b}\\
\dfrac{{\tan A}}{{\tan B}} = \dfrac{{\dfrac{{\sin A}}{{\cos A}}}}{{\dfrac{{\sin B}}{{\cos B}}}} = \dfrac{{\sin A}}{{\sin B}}.\dfrac{{\cos B}}{{\cos A}}\\
= \dfrac{a}{b}.\dfrac{{\dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}}}{{\dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}}}\\
= \dfrac{a}{b}.\dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}:\dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\\
= \dfrac{a}{b}.\dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}.\dfrac{{2bc}}{{{b^2} + {c^2} - {a^2}}}\\
= \dfrac{{{a^2} + {c^2} - {b^2}}}{{{b^2} + {c^2} - {a^2}}}\\
c,\\
A = \dfrac{{\sin 2x + \cos 3x + \sin 6x + \cos 7x}}{{\sin 3x - \sin x}}\\
= \dfrac{{\left( {\sin 2x + \sin 6x} \right) + \left( {\cos 3x + \cos 7x} \right)}}{{\sin 3x - \sin x}}\\
= \dfrac{{2.\sin \dfrac{{2x + 6x}}{2}.cos\dfrac{{6x - 2x}}{2} + 2.\cos \dfrac{{7x + 3x}}{2}.\cos \dfrac{{7x - 3x}}{2}}}{{2.\cos \dfrac{{3x + x}}{2}.sin\dfrac{{3x - x}}{2}}}\\
= \dfrac{{2.\sin 4x.\cos 2x + 2.cos5x.\cos 2x}}{{2\cos 2x.\sin x}}\\
= \dfrac{{\sin 4x + \cos 5x}}{{\sin x}}
\end{array}\)
