Монгол Бодлогын Сан

Заавар: $$\tg2\alpha=\dfrac{2\tg\alpha}{1-\tg^2\alpha}$$ $$\tg(\alpha+\beta)=\dfrac{\tg\alpha+\tg\beta}{1-\tg\alpha\cdot\tg\beta}$$

Бодолт: $A=\tg\Big(2\arctg\dfrac14+\arctg\dfrac{7}{23}\Big)$ илэрхийллийн утгыг бодъё. $$\tg2\arctg\dfrac14=\dfrac{2\tg\arctg\dfrac14}{1-\tg^2\arctg\dfrac14}=\dfrac{2\cdot\dfrac14}{1-\dfrac{1}{4^2}}=\dfrac{8}{15}$$ тул $$\begin{align*} A&=\dfrac{\tg2\arctg\dfrac14+\tg\arctg\dfrac{7}{23}}{1-\tg2\arctg\dfrac14\cdot\tg\arctg\dfrac{7}{23}}\\ &=\dfrac{\dfrac{8}{15}+\dfrac{7}{23}}{1-\dfrac{8}{15}\cdot\dfrac{7}{23}}=\dfrac{8\cdot23+7\cdot15}{15\cdot23-8\cdot7}\\ &=\dfrac{184+105}{345-56}=\dfrac{289}{289}=1 \end{align*}$$ Иймд $$\dfrac{1}{\pi}\Big(2\arctg\dfrac14+\arctg\dfrac{7}{23}\Big)=\dfrac{1}{\pi}\cdot \arctg1=\dfrac{1}{\pi}\cdot \dfrac{\pi}{4}=\dfrac14$$ болно.