Question

Найдите tg a,если известно,что cosa=2/корень из 5,0<a<pi/2.

Answer 1

$tgalpha-?;\ cosalpha=frac{2}{sqrt5};\ 0<alpha<frac{pi}{2};\ 0<alpha<frac{pi}{2}==>cosalpha>0; sinalpha>0; tgalpha>0;\ 1+tg^2alpha=1+frac{sin^2alpha}{cos^2alpha}=frac{cos^2alpha+sin^2alpha}{cos^2alpha}=frac{1}{cos^2alpha};\ tg^2alpha=frac{1}{cos^2alpha}-1=frac{1-cos^2alpha}{cos^2alpha};$
$tgalpha=+sqrt{frac{1-cos^2alpha}{cos^2alpha}}=+sqrt{frac{1-left(frac{2}{sqrt5}right)^2}{left(frac{2}{sqrt5}right)}}=+sqrt{frac{1-frac{4}{5}}{frac{4}{5}}}=\ =+sqrt{frac{frac{5-4}{5}}{frac{4}{5}}}=+sqrt{frac{frac15}{frac45}}=+sqrt{frac{1cdot5}{5cdot4}}=+sqrtfrac14=+frac12=frac12;\ \ \ \ \ tgalpha=frac12$