√2sin45°-cos30°sin60°+ctg45°tg135°-tg0

Question

√2sin45°-cos30°sin60°+ctg45°tg135°-tg0

Answer 1

√2*√2/2 - √3/2*√3/2 +ctg45°*ctg(180°-45°)-0=
=1-1,5+ctg45°*(-tg45°)=-0,5-1=-1,5