Монгол Бодлогын Сан

Заавар: $$\cos\alpha\cos\beta-\sin\alpha\sin\beta=\cos(\alpha+\beta)$$ ба $$2\sin\alpha\cos\alpha=\sin2\alpha$$

Бодолт: $$\dfrac{1}{\sin10^\circ}-\dfrac{\sqrt3}{\cos10^\circ}=\dfrac{4\big(\frac12\cdot\cos10^\circ-\frac{\sqrt{3}}{2}\cdot\sin10^\circ\big)}{2\sin10^\circ\cos10^\circ}=$$ $$=\dfrac{4(\cos60^\circ\cdot\cos10^\circ-\sin60^\circ\cdot\sin10^\circ)}{2\sin10^\circ\cos10^\circ}=\dfrac{4\cos(60^\circ+10^\circ)}{\sin20^\circ}=4$$