Упростите выражение: 2cos2a*tga/tg2a

Ответ:

$\dfrac{\text{cos}^{2}2\alpha}{\text{cos}^{2}\alpha}$

Пошаговое объяснение:

$2 \ \text{cos} \ 2\alpha \ \cdotp \dfrac{\text{tg} \ \alpha}{\text{tg} \ 2 \alpha} = 2 \ \text{cos} \ 2\alpha \ \cdotp \dfrac{\text{tg} \ \alpha}{\dfrac{2 \ \text{tg} \ \alpha}{1 - \text{tg}^{2} \alpha}} = 2 \ \text{cos} \ 2\alpha \ \cdotp \dfrac{1 - \text{tg}^{2} \alpha}{2} = \\\\= \text{cos} \ 2\alpha \ \cdotp \bigg(1 - \dfrac{\text{sin}^{2}\alpha}{\text{cos}^{2}\alpha} \bigg) = \text{cos} \ 2\alpha \ \cdotp \bigg(\dfrac{\text{cos}^{2} \alpha - \text{sin}^{2}\alpha}{\text{cos}^{2}\alpha} \bigg) =$

$= \text{cos} \ 2\alpha \ \cdotp \dfrac{\text{cos} \ 2\alpha}{\text{cos}^{2}\alpha} = \dfrac{\text{cos}^{2}2\alpha}{\text{cos}^{2}\alpha}$