Question

(cos(-135)+sin(-120)-cos(-120))/(tg^2(690)+ctg^2(-30))

Answer 1

=cos(180°-45°)-sin(180°-60°)-cos(180°-60°)/
((tg720°-30°))²+ctg²30°=
(-cos45°-sin60°+cos60°)/(tg²30°+ctg²30°)
=(-√2/2-√3/2+1/2)/(√3/3)²+(√3)²)=
(√3-√2+1)/2:(1/3+3)=
(√3-√2+1)/2*3/10=3/20*(√3-√2+1)