Đáp án:
$\begin{array}{l}
a) - \pi < a < - \dfrac{\pi }{2}\\
\Rightarrow \cos a < 0\\
\Rightarrow \cos a = - \sqrt {1 - {{\sin }^2}a} = - \dfrac{{2\sqrt 2 }}{3}\\
\Rightarrow \sin 2a = 2.\sin a.\cos a = \dfrac{{4\sqrt 2 }}{9}\\
\Rightarrow \tan a = \dfrac{{\sin a}}{{{\mathop{\rm cosa}\nolimits} }} = \dfrac{{\sqrt 2 }}{4}\\
b)\dfrac{{\sin \left( {a - b} \right)}}{{\cos a.\cos b}} + \dfrac{{\sin \left( {b - c} \right)}}{{\cos b.{\mathop{\rm cosc}\nolimits} }}\\
= \dfrac{{\sin a.\cos b - \sin b.\cos a}}{{\cos a.cosb}} + \dfrac{{\sin b.\cos c - \sin c.\cos b}}{{\cos b.{\mathop{\rm cosc}\nolimits} }}\\
= \dfrac{{\sin a}}{{\cos a}} - \dfrac{{\sin b}}{{\cos b}} + \dfrac{{\sin b}}{{\cos b}} - \dfrac{{\sin c}}{{{\mathop{\rm cosc}\nolimits} }}\\
= \dfrac{{\sin a}}{{\cos a}} - \dfrac{{{\mathop{\rm sinc}\nolimits} }}{{\cos c}}\\
= \dfrac{{\sin a.\cos c - {\mathop{\rm sinc}\nolimits} .cosa}}{{\cos a.{\mathop{\rm cosc}\nolimits} }}\\
= \dfrac{{\sin \left( {a - c} \right)}}{{\cos a.{\mathop{\rm cosc}\nolimits} }}\\
Vậy\,\dfrac{{\sin \left( {a - b} \right)}}{{\cos a.\cos b}} + \dfrac{{\sin \left( {b - c} \right)}}{{\cos b.{\mathop{\rm cosc}\nolimits} }} = \dfrac{{\sin \left( {a - c} \right)}}{{\cos a.{\mathop{\rm cosc}\nolimits} }}
\end{array}$
