Монгол Бодлогын Сан

Заавар: $\dfrac{1}{\sin^2\alpha}=\dfrac{\sin^2\alpha+\cos^2\alpha}{\sin^2\alpha}=1+\ctg^2\alpha$ ашиглан бод.

Бодолт: \begin{align*} \sqrt{2\ctg\alpha+\dfrac{1}{\sin^2\alpha}}&=\sqrt{1+2\ctg\alpha+\ctg^2\alpha}\\ &=\sqrt{(1+\ctg\alpha)^2}=|1+\ctg\alpha| & &\color{red}{\leftarrow\sqrt{a^2}=|a|} \\ &=1+\dfrac{1}{2}=1.5 & &\color{red}{\leftarrow\ctg\alpha=\dfrac{1}{\tg\alpha}} \end{align*}