Question

2 cos 30 ° ctg 60 °

2 cos 60° + tg 45 °

cos 45° -2 sin 30°

Answer 1

$2cos 30^o ctg 60^o=2*frac{sqrt{3}}{2}*frac{1}{sqrt{3}}=1;$

$2cos 60^o+tg 45^o=2*frac{1}{2}+1=1+1=2;$

$cos 45^o-2sin 30^o=frac{sqrt{2}}{2}-2*frac{1}{2}=frac{sqrt{2}-1}{2}$