Монгол Бодлогын Сан

Заавар: $\tg\alpha=\dfrac{\sin\alpha}{\cos\alpha}$, $\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$ ба $\sin2\alpha=2\sin\alpha\cos\alpha$ томьёонуудыг ашиглан бод.

Бодолт: $$\dfrac{\tg3\alpha+\tg\alpha}{\tg3\alpha-\tg\alpha}=\dfrac{\dfrac{\sin3\alpha}{\cos3\alpha}+\dfrac{\sin\alpha}{\cos\alpha}}{\dfrac{\sin3\alpha}{\cos3\alpha}-\dfrac{\sin\alpha}{\cos\alpha}}=\dfrac{\dfrac{\sin3\alpha\cos\alpha+\cos3\alpha\sin\alpha}{\cos3\alpha\cos\alpha}}{\dfrac{\sin3\alpha\cos\alpha-\cos3\alpha\sin\alpha}{\cos3\alpha\cos\alpha}}=$$ $$=\dfrac{\sin3\alpha\cos\alpha+\cos3\alpha\sin\alpha}{\sin3\alpha\cos\alpha-\cos3\alpha\sin\alpha}=\dfrac{\sin4\alpha}{\sin2\alpha}=\dfrac{2\sin2\alpha\cos2\alpha}{\sin2\alpha}=2\cos2\alpha$$