Question

помогите, умоляю . срочно

Answer 1

α ∈ (π , 2π ) , значит Sinα < 0

$1+Ctg^{2}\alpha=\frac{1}{Sin^{2}\alpha}\\\\Sin^{2}\alpha=\frac{1}{1+Ctg^{2}\alpha}=\frac{1}{1+(2\sqrt{6})^{2}}=\frac{1}{1+24}=\frac{1}{25}\\\\Sin^{2}\alpha=\frac{1}{25}\\\\Sin\alpha=-\frac{1}{5}=-0,2$