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

\cos\alpha>0;\ \sin\alpha>0;\ tg\alpha>0;\\ 1+tg^2\alpha=1+\frac{\sin^2\alpha}{\cos^2\alpha}=\frac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}=\frac{1}{\cos^2\alpha};\\ tg^2\alpha=\frac{1}{\cos^2\alpha}-1=\frac{1-\cos^2\alpha}{\cos^2\alpha};\\" alt="tg\alpha-?;\\ \cos\alpha=\frac{2}{\sqrt5};\\ 0<\alpha<\frac{\pi}{2};\\ 0<\alpha<\frac{\pi}{2}==>\cos\alpha>0;\ \sin\alpha>0;\ tg\alpha>0;\\ 1+tg^2\alpha=1+\frac{\sin^2\alpha}{\cos^2\alpha}=\frac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}=\frac{1}{\cos^2\alpha};\\ tg^2\alpha=\frac{1}{\cos^2\alpha}-1=\frac{1-\cos^2\alpha}{\cos^2\alpha};\\" align="absmiddle" class="latex-formula">
$tg\alpha=+\sqrt{\frac{1-\cos^2\alpha}{\cos^2\alpha}}=+\sqrt{\frac{1-\left(\frac{2}{\sqrt5}\right)^2}{\left(\frac{2}{\sqrt5}\right)}}=+\sqrt{\frac{1-\frac{4}{5}}{\frac{4}{5}}}=\\ =+\sqrt{\frac{\frac{5-4}{5}}{\frac{4}{5}}}=+\sqrt{\frac{\frac15}{\frac45}}=+\sqrt{\frac{1\cdot5}{5\cdot4}}=+\sqrt\frac14=+\frac12=\frac12;\\ \\ \\ \\ \\ tg\alpha=\frac12$