Question

$sin ^{2} 150$+$cos^{2} 120$+$tg^{2} 240$-$ctg^{2} 135$

Answer 1

 

     $sin^2150+cos^2120+tg^2240-ctg^2135=\ sin^2(30*5)+cos^2(60*2)+tg^2(120*2)-ctg^2(3*45)=\ frac{1}{2}^2+(-frac{1}{2})^2+(sqrt{3})^2-(-1)^2=\ frac{1}{4}+frac{1}{4}+3-1=frac{5}{2}$