Giải phương trình.
$\dfrac{\sin(a-b)}{\cos a.\cos b}$
$=\dfrac{\sin a.\cos b -\cos a.\sin b}{\cos a.\cos b}$
$=\dfrac{\sin a.\cos b}{\cos a.\cos b}-\dfrac{\cos a.\sin b}{\cos a.\cos b}$
$=\tan a-\tan b= \tan a+\tan b$
$\Leftrightarrow 2\tan b=0$
$\Leftrightarrow \tan b=0$ (*)
ĐK: $a, b\neq \dfrac{\pi}{2}+k\pi$
(*) $\Leftrightarrow b=k\pi$ (TM)
$\Rightarrow a\in \mathbb{R}$ \ $\{\dfrac{\pi}{2}+k\pi\}$
