Cos( arctg 1/2- arctg 2)

\cos x=\frac{1}{\sqrt{1+tg^2x}};\\ \sin^2x=1-\cos^2x=1-\frac{1}{tg^2x+1}=\frac{tg^2x+1-1}{tg^2x+1}=\frac{tg^2x}{tg^2x+1};\\ \sin x=\frac{tgx}{\sqrt{tg^2x+1}}|\\ " alt="\cos(arctg\frac{1}{2}-arctg2)=\\ \cos(arctg\frac{1}{2})\cos(arctg2)+\sin(arctg\frac{1}{2})\sin(arctg2)=\\ =|tg^2x+1=\frac{1}{\cos^2x}==>\cos x=\frac{1}{\sqrt{1+tg^2x}};\\ \sin^2x=1-\cos^2x=1-\frac{1}{tg^2x+1}=\frac{tg^2x+1-1}{tg^2x+1}=\frac{tg^2x}{tg^2x+1};\\ \sin x=\frac{tgx}{\sqrt{tg^2x+1}}|\\ " align="absmiddle" class="latex-formula">
$=\frac{1}{\sqrt{1+tg^2(arctg\frac{1}{2})}}\cdot\frac{1}{\sqrt{1+tg^2(arctg2)}}+\frac{tg(arctg\frac{1}{2})}{\sqrt{1+tg^2(arctg\frac{1}{2})}}\cdot\frac{tg(arctg2)}{\sqrt{1+tg^2(arctg2)}}=\\ =\frac{1}{\sqrt{1+(\frac{1}{2})^2}}\cdot\frac{1}{\sqrt{1+2^2}}+\frac{\frac{1}{2}}{\sqrt{1+(\frac{1}{2})^2}}\cdot\frac{2}{\sqrt{1+2^2}}=\\ =\frac{1}{\sqrt{(1+\frac{1}{4})(1+4)}}+\frac{1}{\sqrt{(1+\frac{1}{4})(1+4)}}=\\ =\frac{2}{\sqrt{(1+\frac{1}{4})(1+4)}}=\frac{2}{\frac{1}{2}\sqrt{(1+4)(1+4)}}=\\$
$=\frac{4}{\sqrt{(1+4)^2}}=\frac{4}{1+4}=\frac{4}{5}$