Question

Qual o valor de:
           cos.150°+sen300°-tg225°-cos90°=

Answer 1

$cos.150^o = -cos30^o =-\frac{\sqrt3}{2} \\ sen 300^o = -sen 60^o -\frac{\sqrt3}{2}\\ tg 225^o = tg 45^o = 1 \\ cos 90^o=0 \\ \\ cos.150^o+sen300^o-tg225^o-cos90^o=-\frac{\sqrt3}{2}-\frac{\sqrt3}{2}-1-0= \\ \\ \boxed{-\sqrt3-1}$