💖💖💖恒等变换

$\mathrm{在}\mathrm{△}ABC\mathrm{中}，A+B+C=\pi$
$\cos C=-\cos (A+B)$
$=-\cos A\cos B+\sin A\sin B$
$A\in (0,\pi )$
$\cos B>0,B\in (0,\frac{\pi }{2})$
$\sin B=\frac{12}{13}$
$\cos A=\sqrt{1-{\sin }^{2}A}=\pm \frac{4}{5}$
$\because \sin B>\sin A$
$\therefore A<B$
$\therefore \cos A>0$
$\cos A=\frac{4}{5}$
$\cos C=\frac{16}{65}$

$\mathrm{原}\mathrm{式}=\sin {40}^{\circ }(\frac{\sin {10}^{\circ }}{\cos {10}^{\circ }}-\sqrt{3})$
$=\sin {40}^{\circ }\frac{\sin {10}^{\circ }-\sqrt{3}\cos {10}^{\circ }}{\cos {10}^{\circ }}$
$=\sin {40}^{\circ }\frac{2(\frac{1}{2}\sin {10}^{\circ }-\frac{\sqrt{3}}{2}\cos {10}^{\circ })}{\cos {10}^{\circ }}$
$=\sin {40}^{\circ }\frac{2\sin ({10}^{\circ }-{60}^{\circ })}{\cos {10}^{\circ }}$
$=\frac{-2\sin {40}^{\circ }\sin {50}^{\circ }}{\cos {10}^{\circ }}$
$=\frac{-2\sin {40}^{\circ }\cos {40}^{\circ }}{\cos {10}^{\circ }}$
$=-\frac{\sin {80}^{\circ }}{\cos {10}^{\circ }}=-1$

$\mathrm{原}\mathrm{式}=\sin {40}^{\circ }(\tan {10}^{\circ }-\tan {60}^{\circ })$
$=\sin {40}^{\circ }\tan ({10}^{\circ }-{60}^{\circ })(1+\tan {10}^{\circ }\tan {60}^{\circ })$
$=-\cos {50}^{\circ }\tan {50}^{\circ }\frac{\cos {10}^{\circ }\cos {60}^{\circ }+\sin {10}^{\circ }\sin {50}^{\circ }}{\cos {10}^{\circ }\cos {60}^{\circ }}$
$=-\sin {50}^{\circ }\frac{\cos {50}^{\circ }}{\cos {10}^{\circ }\cos {60}^{\circ }}$
$=-\frac{2\sin {50}^{\circ }\cos {50}^{\circ }}{\cos {10}^{\circ }}=-\frac{\sin 100\circ }{\cos {10}^{\circ }}=-\frac{\cos {10}^{\circ }}{\cos {10}^{\circ }}=-1$