Question

31 номер тригонометрия

Answer 1

$cos alpha cdot(1+tg alpha +cos^{-1} alpha )(1+tg alpha -cos^{-1} alpha )=\ \ \ =cos alpha cdot ((1+tg alpha )^2-(cos^{-1} alpha )^{2})=\ \ \ =cos alpha cdot(1+2tg alpha +tg^2 alpha - cos^{-2} alpha )=\ \ \ =cos alpha cdot(2tg alpha + frac{1}{cos^2 alpha } - frac{1}{cos^2 alpha } )=2cos alpha cdot tg alpha =2sin alpha$