Монгол Бодлогын Сан

Заавар:

Бодолт: $$\dfrac{\cos2\alpha}{1+\cos2\alpha}=\dfrac{\tg\alpha}{\tg2\alpha}\Leftrightarrow\cos2\alpha=\dfrac{\tg\alpha}{\tg2\alpha}\cdot(1+\cos 2\alpha)$$ $$\begin{align*} \dfrac{\tg\alpha}{\tg2\alpha}\cdot(1+\cos 2\alpha)&=\dfrac{\tg\alpha}{\tg2\alpha}\cdot2\cos^2\alpha & & \color{red}{\leftarrow 1+\cos2\alpha=2\cos^2\alpha}\\ &=\dfrac{\sin\alpha/\cos\alpha}{\sin2\alpha/\cos2\alpha}\cdot2\cos^2\alpha & & \color{red}{\leftarrow \tg\alpha=\dfrac{\sin\alpha}{\cos\alpha},\ \tg2\alpha=\dfrac{\sin2\alpha}{\cos2\alpha}}\\ &=\dfrac{\sin\alpha\cdot\cos2\alpha}{\sin2\alpha\cdot\cos\alpha}\cdot2\cos^2\alpha\\ &=\dfrac{(2\sin\alpha\cos\alpha)\cdot\cos2\alpha}{\sin2\alpha}=\cos2\alpha & & \color{red}{\leftarrow2\sin\alpha\cdot\cos\alpha=\sin2\alpha} \end{align*}$$