Đáp án:
$\begin{array}{l}
+ )0 < a < \dfrac{\pi }{2}\\
\Rightarrow \cos a > 0\\
\Rightarrow \cos a = \sqrt {1 - {{\sin }^2}a} = \dfrac{1}{3}\\
\Rightarrow \tan a = 2\sqrt 2 ;{\mathop{\rm cota}\nolimits} = \dfrac{{\sqrt 2 }}{4}\\
+ \dfrac{{3\pi }}{2} < b < 2\pi \\
\Rightarrow {\mathop{\rm sinb}\nolimits} < 0;cosb > 0\\
Do:\dfrac{1}{{{{\sin }^2}b}} = {\cot ^2}b + 1 = \dfrac{{25}}{{16}}\\
\Rightarrow \sin b = - \dfrac{4}{5}\\
\Rightarrow \cos b = \dfrac{3}{5}\\
\Rightarrow \tan b = - \dfrac{4}{3}\\
+ )\sin \left( {a - b} \right) = \sin a.\sin b - \cos a.\cos b\\
= - \dfrac{{3 + 8\sqrt 2 }}{{15}}\\
+ )cos\left( {a + b} \right)\\
= \cos a.sinb - \sin a.\cos b\\
= - \dfrac{{4 + 6\sqrt 2 }}{{15}}\\
+ )\tan \left( {a - b} \right)\\
= \dfrac{{\tan a - \tan b}}{{1 + \tan a.\tan b}} = - 1,5
\end{array}$
