Монгол Бодлогын Сан

Заавар: $$\cos^2\alpha+\sin^2\alpha=1\Rightarrow1+\dfrac{1}{\ctg^2\alpha}=\dfrac{1}{\cos^2\alpha}\Rightarrow \cos^2\alpha=\dfrac{\ctg^2\alpha}{1+\ctg^2\alpha}$$

Бодолт: $$\cos^2\alpha=\dfrac{\ctg^2\alpha}{1+\ctg^2\alpha}=\dfrac{\Big(\dfrac{5}{12}\Big)^2}{1+\Big(\dfrac{5}{12}\Big)^2}=\dfrac{25}{169}$$ ба $\pi<\alpha<\dfrac{3\pi}{2}$ тул $\cos\alpha<0$ байна. Иймд $\cos\alpha=-\sqrt{\dfrac{25}{169}}=-\dfrac{5}{13}$ байна.