Question Number 109625 by qwerty111 last updated on 24/Aug/20
$$\frac{\mathrm{sin}\:\mathrm{2}\boldsymbol{\alpha}+\mathrm{2sin}\:\boldsymbol{\alpha}\centerdot\mathrm{cos}\:\mathrm{2}\boldsymbol{\alpha}}{\mathrm{1}+\mathrm{cos}\:\boldsymbol{\alpha}+\mathrm{cos2}\:\boldsymbol{\alpha}+\mathrm{cos3}\:\boldsymbol{\alpha}} \\ $$
Commented by nimnim last updated on 24/Aug/20
$${tan}\alpha \\ $$
Commented by kaivan.ahmadi last updated on 24/Aug/20
$$\frac{\mathrm{2}{sin}\alpha{cos}\alpha+\mathrm{2}{sin}\alpha\left(\mathrm{2}{cos}^{\mathrm{2}} \alpha−\mathrm{1}\right)}{\mathrm{1}+{cos}\alpha+\left(\mathrm{2}{cos}^{\mathrm{2}} \alpha−\mathrm{1}\right)+\left(\mathrm{4}{cos}^{\mathrm{3}} \alpha−\mathrm{3}{cos}\alpha\right)}= \\ $$$$\frac{\mathrm{2}{sin}\alpha\left(\mathrm{2}{cos}^{\mathrm{2}} \alpha+{cos}\alpha−\mathrm{1}\right)}{\mathrm{4}{cos}^{\mathrm{3}} \alpha+\mathrm{2}{cos}^{\mathrm{2}} \alpha−\mathrm{2}{cos}\alpha}=\frac{\mathrm{2}{sin}\alpha\left(\mathrm{2}{cos}^{\mathrm{2}} \alpha+{cos}\alpha−\mathrm{1}\right)}{\mathrm{2}{cos}\alpha\left(\mathrm{2}{cos}^{\mathrm{2}} \alpha+{cos}\alpha−\mathrm{1}\right)}=\frac{{sin}\alpha}{{cos}\alpha} \\ $$$${tg}\alpha\: \\ $$