Монгол Бодлогын Сан

Заавар: $$\sin\alpha\sin\beta=-\dfrac{1}{2}(\cos(\alpha+\beta)-\cos(\alpha-\beta))$$ $$\cos\alpha\cos\beta=\dfrac{1}{2}(\cos(\alpha+\beta)+\cos(\alpha-\beta))$$

Бодолт: $$\tg\alpha\cdot \tg\beta=\dfrac{\sin\alpha}{\cos\alpha}\cdot\dfrac{\sin\beta}{\cos\beta}=\dfrac{\sin\alpha\sin\beta}{\cos\alpha\cos\beta}=$$ $$=\dfrac{-\cancel{\dfrac{1}{2}}(\cos(\alpha+\beta)-\cos(\alpha-\beta))}{\cancel{\dfrac{1}{2}}(\cos(\alpha+\beta)+\cos(\alpha-\beta))}=-\dfrac{\dfrac{1}{3}-\dfrac15}{\dfrac{1}{3}+\dfrac15}=-\dfrac{1}{4}$$