Question

2$sin^{2}$a+$sqrt{2}$cosa+tga, ctga=1

Answer 1

$ctg(a)=1\a=arcctg1\a=frac{pi}{4}$


$(sina)^2+sqrt{2}cosa+tga=(sinfrac{pi}{4})^2+sqrt{2}*cosfrac{pi}{4}+tgfrac{pi}{4}=\=(frac{sqrt{2}}{2})^2+sqrt{2}*frac{sqrt{2}}{2}+1=frac{1}{2}+1+1=2.5$