Question

sin(a+3)+sin(a-3)/cos(a+b)+cos(a-b)

Answer 1

$frac{sin(a+3)+sin(a-3)}{cos(a+b)+cos(a-b)}=frac{sinacos3+cosasin3+sinacos3-cosasin3}{cosacosb-sinasinb+cosacosb+sinasinb}=frac{2sinacos3}{2cosacosb}=$
$=tga*frac{cos3}{cosb}$